JEE (Main) - PHYSICS Module-3 (Class XI) @® Aakash Medical|IIT-JEE| Foundations5. Work, Energy and Power.. .01 - 76 6. ‘System of Particles and Rotational Motion .. 77 — 190 Answers... 191 - 194Cuapter 5 Work, Energ INTRODUCTION " We use the term ‘work’ in every day conversation to mean many Introduction different things. We talk about going to work, doing homework, The Scalar Product or working in class. Physicists mean something very specific when Dot Product Between Two they talk about work. In physics we usc the term work to describe Vectors the process of transferring energy from object or system to another or converting energy from one form to another. In this chapter, Work Done by a Force we shall study notion of work and energy and relation between + Energy work and energy by the help of work energy theorem. In order to learn concept of work we first need to learn scalar product of two vectors. + Kinetic Energy + The Concept of Potential Energy THE SCALAR PRODUCT OR DOT PRODUCT BETWEEN + Gravitational Potential TWO VECTORS Energy / * There are two ways of multiplying vectors (i) scalar product Elastic Potential Energy of (li) vector product. We shall study the latter in next chapter. The Spring scalar product or dot product of any two vectors A and B, The Work-Energy Theorem denoted as A-B (read as Adot B) is defined as A-B=ABcos0 Where A & B are magnitudes of vectors AandB respectively and @ is the smaller angle between them. Dot product is callled scalar product as A, B and cos@ are scalars. Both vectors have a direction but their scalar product does not have a direction. Power + The Conservation of Mechanical Energy + Motion in a Vertical Circle + Collisions + Some Important Definitions + Formulae Chart + Quick Recap Geometrical interpretation B00 $ © Fig.:4 Let AandB be represented by OG andOR respectively. Let ZQOR be 6. In figure (b) draw RS 1 OQ. OS = Bcos@ is the projection of B.onto A, . In figure (c), draw QS_L OR OS = Acos@ is the projection of A onto B. A (Bcos@) = B(Acos6)SESE Work, Energy and Power Board, JEE (Main) & State Competitive Exams AB is the product of magnitude of A and component of B along A (or projection of B onto A). Altematively, itis the product of magnitude of B andthe componentof A along & (or projection of A onto B). . Properties () Dot product is commutative ie, |A-B=B-A (ie., dot product is commutative) (i) Dot product is distributive ie., [A (B+ 6)=A-B+A:C (ji) Dot product of a vector with itself gives square of its magnitude ie, [A-A=AAcos0=A* (vy [A-@8)=24-B) where 2 is a real number. @) [RA08)=6A-6) where 2 is a real number. Special cases : Two vectors are perpendicular (i.e., @ = 90°) A. B= ABcos90° =0 Case4: For unit vectors 7-j=j-k=k-i=0 Gase-ll : - Two vectors are parallel (i.e., 0 = 0°) A-B = ABcos0 = AB(1)= AB For unit vectors fia] j=kk=A * Gase-il: Two vectors are antiparallel (/ A. B= ABcos 180 =- AB, Understanding dot product in cartesian coordinates i+ Ayj+Azk & B= Bi +B,j+ Bak (Ad +A,j+ Ack): (xi + By] + BA) =AB,+ ABy+ AB B+ A,B, + AB, If the magnitude of two vectors are 4 and 6 and the magnitude of their scalar product is 12,2, what is the angle between the vectors? 4,B=6,A-B= 1242 B= ABcos0 cos 2 AB 422 _ 4 AB 4x6 2 cos® = cos45" = 45° Solution : [EQEEEM Fina the angle between force F= (51 + 4] +5k) unit and displacement d= (3 + 4] Also find the projection of F ond. :ee Board, JEE (Main) & State Competitive Exams_ Work, Energy and Power aes Fd, + Fydy+ Fd, 5(3) + 4(4) + 5(-3) = 16 unit ~ Solution : F SFL+ Fy +R? =8°+4?+5? = 66unit d-d=d? = d?+d7+?=3°+4? + (-3)?=34 unit F-d=Fdcoso _ exetiaD 16 80" Fg ~ eee avaTO 0 = cos“(0.34) . Projection of F ond = Fcos6 = 466 x 0.34 = 2.76 WORK DONE BY A FORCE . “The work done by a force (F) on a body is defined as we [Far When dr is the displacement of the point of application of force in a very small time interval. Case: if F is constant. fa we fear W=F-Ar Ar is the total displacement of the particle.Work, Energy and Power Board, JEE (Main) & State Competitive Exams A particle of mass m is released in a smooth hemispherical bow! from shown position A. Find work doné by gravity as it reaches the lowest point B. Solution: Displacement of the particle d=Ri2 Weight will remain constant throughout the process. Work done by the weight x Aw= mgdcos =moR Oe => Aw=mgR A body is subjected to a constant force given by F(N)=-i+2)+3k. What is the work done by this force in moving the body through a distance of 4 m along the z-axis and then 3 m along the y-axis? Solution: The formula to be used is W=F.S For first part of the displacement, we have ‘S(m)=4k So, work done is given as We F.S = (i +2j+3h).(4K) = 12 Nm= 123 For second part of the displacement, we have S(m)=3 So, work done is given as W=FS = (-1+2]+3k).(3)) =6N-m=6J Total work done is thus given by W= 12+6= 18 J The distance x moved by a body of mass 0.5 kg under the action of a force varies with tim tas x (m) = 38 + 4t+ 5. ; Here, tis expressed in second. What is the work done by the force in first 2 seconds? Solution: . From the displacement equation, one can determine the expression for force acting on th object. It has been calculated below x= 32+ att 5 Differentiating once, w.rt. time, we get dx a td or v=6t+4 Differentiating again, w.r-t. time, we get we ae => a=6 mis? So, the force is F= ma=0.5x6=3NBoard, JEE-(Main) & State Competitive Exams Work, Energy and Power Jaman Case-ll : ‘The body moves in a straight line and positive sign in force and velocity for the interval 0:2: shows that the angle between force and displacement is 0°. So, work done is W/= FS cos0”. The displacement S = x,~ xp [3(2)? + 4(2) + 5] — [3(0) + 4(0) + 5] 124+8+5-5=20m So, work done is W=3 x 20=60J Ifforce F is variable. In daily life, a constant force is rare e.g., if we want to calculate work done by an engine of a car, force exerted by engine is a variable force (car accelerates, travels with constant speed for some time, brakes are applied ...it comes to rest ... again accelerates ...) Fig. :2 Fig. 2 shows plot of a varying force in one dimension. If displacement Ax is small, force F(x) can be regarded as approximately constant and the work done AW = F(x)Ax. This is shown in Fig. 2, AW is the work done corresponding to small displacement ‘Ax over which force is regarded constant. If we add successive rectangular areas in Fig. 2. Total work done, W= >7 F(x)dx where the summation is from initial positic \n x to the final position x, ef Asan example, if we are talking about >) F(x)Ax Itis = Fax + Fax + Fax + F,Ax = If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 3 Fa) OX x % Fig. : 3 ; ‘Adding the areas.of all the rectangles we find that for Ax —> 0, the area under the curve is exactly equal to the work done by F(x). w= lim SF o)ax= J Fea where ‘lim’ stands for the limit of the sum when Ax tends to zero. Thus, for a varying fore the work done can be expressed .as a definite integral of force over displacement.TE Work, Energy and Power Board, JEE (Main) & State Competitive Exams If force F=(3xi+y’})\N is acting on a body and body moves from (1 m, 2m, 1 m) to (3m, 3m, 8m), then find the work done due to the force. Solution: — W=[ Fde+f Rydy+f Fede a) = [ axax+ f y'dy+0 [As F:=0) ia e144 19 55. 3 1 =F M+ gR7-81= 124-5 =F) Calculation of Work Done using Graphs Work done W= | F.dF, work done by a force can be calculated by calculating the area under the force displacement curve, Force versus displacement curve is shown in the diagram: Find the work done by the force at the end of the displacements. FM s(n) (10m (i) 20m (ii) 30m Solution: _ As we know that work done is area under F — s curve. (i) Work done at the end of 10 m W =A, (Area from 0 to 10 m) = Fx10x105 =504 (i) Work done at the end of 20 m W =A, +A, (Area from 0 to 20 m) =50+50=100J (ii) Work done at the end of 30. m W =A,+A,+A; (Area from 0 to 30 m) = 50 + 50-50 (A) =~ 50) = 50J The force exerted on an object is F = F,(-2-~ 1). Find the work done in moving the object from x = 0 to x = 3x» % (a) by plotting F(x) and using the area under the curve, and (6) by evaluating the integral analytically. ‘Solution : (a) The graph shows F as a function of x. The work is negative as object moves from x = 0 to x = x)and positive as it moves from x = Xp to x = 3X9. ‘Since the area of a — triangle is ($) base), (altitude), the work done for the movement from x = 0 to. wia( \(x0)(Fo) and the work-done for the movement from x = X, to x = 3% is“Board, JEE (Main) & State Competitive Exams Work, Energy and Power eee ($)%0-x9(2F)= 2108. a 2F, : The total work done is the sum of them, 3 which is Fe (F).- (b) The integral for the work is ; Ky 2k 3% - we fA (2 see 5 [2a =(S)oaneo ENERGY Energy is a physical quantity which enables a system to perform work. There are so many form of energy like heat energy, sound energy, nuclear energy which will be discussed in their respective topics. Here we will discuss mechanical energy. Mechanical Energy Mechanical energy is the energy that is possessed by an object due to its motion or due to its position mechanical energy of an object is the sum of its kinetic and potential energy. Mechanical energy = KE + PE KINETIC ENERGY ‘As discussed eariier, kinetic energy K of an object of mass mis given as, Itis a scalar quantity. It is a measure of the work an object can do because of its motion. Sailing ships use KE of the wind. KE of a fast-flowing stream has been used for grinding corn. Relation between KE and linear momentum Let mass of body be m and velocity of body be v. p= mv= linear momentum Kt Graphical Representation The above relation between momentum and kinetic energy can be represented by following graphs. The graph-1 gives a variation between K and p for a given mass, while the graph-2 gives a variation between K and m for a given momentum.Board, JEE (Main) & State Competitive Exams SEE! Work, Eneray and Power Graph-1 Graph-2 Ki KA rectangular 7 KL hyperbola parabola (y= cx’) (x=0) “Ashooter fires a bullet of mass 100 g with a speed of 100 ms-* on a soft plywood of thickness. ‘tem, The bullet emerges with 10% ofits initial KE. Find the emergent speed of the bullet. it od gy?= tx 1. = Solution: Initial KE of the bullet = mu? => sop % 100 x 100 = 5003 Final KE of the bullet = < x 500 = 50 Let emergent speed of the bullet be v, Swe = v= PRE 4000 THE CONCEPT OF POTENTIAL ENERGY Conservative and Non-conservative Forces in the previous section, work done by various forces have been discussed. The various forces can be classified into two types depending on the nature of work done by them. 3 1. Conservative Forces : When a body moves under the action of a conservative force, the work § j Gone by the conservative force depends on the initial and final position of the body and it sq” independent of the path followed by the body. Consider the following situation. A particle moves |) 4. under the action of a force F from A to B via two different paths 1 and 2. : n B , & 1.6ms* pane? A For path 1, W, represents the work done and for path 2, W. represents the work done If F is a conservative force, W, = Ws. Now, if the particles moves from A to B via path 1 and returns to A via path 2, total wor done on the particle by the force is zero. This means that while moving from A to B, the for supplied some energy to the pariicle and during its return journey from any path, from 6 to A iLbkes away the same amount of energy from the particle. Or, for a round trip or closed path there ig no loss oF gain of energy by the particle, This is why, the force is called conservativ force. The examples of conservative forces are gravitational force, spring force and electrostati force.Board, JEE (Main) & State Competitive Exams Work, Energy and Power ees 2. Non-Conservative Force : When the work done by a force during the movement of a particle from A to Bis different for different paths, the force is said to be non-conservative. In this situation, if a particle moves from A to B via a path 1 and retumns from B to A via a path 2, the total work done will not be zero. That is, Wass + Ware # 0 or [Wao] # [Weasl- Usually the total work done by the force during the round trip is, east 8 negative. If it is so, then the force is called dissipative force as it dissipates the energy of the body. The most common example of dissipative force is force of friction. It dissipates the energy of the particle in the form of heat. A Potential Energy ' When a body moves under the action of a conservative force, it is seen that the work done by the force depends only on initial and final position of the body and for a round irip, work done by the force fe is zero. Let a body move from A to B via path 1 and retum to A via path 2 under the action of a conservative force F. At A, the kinetic energy of the body is K, and at B, kinetic energy of oo 8 the body is K;,. Let K, < K,. When body moves from A to 8, its kinetic k) energy (K,) decreases. It means, negative work has been done by the force F As it moves back to A, same amount of positive work is done by the A, pa? force F such that the body again acquires the kinetic energy K,. itis (K) clear that, as body moves from A to B, the force F takes away energy from it and as the body moves back from B to A, the force F gives back the energy to it. Potential energy is not only a property of conservative force, in fact and it should bé clearly understood that the potential energy is associated with the configuration of the system constituting the particle and the arrangement which produces the conservative force. Mathematical Expression for Potential Energy When a body moves under the action of a conservative force F through a small displacement dr, the work done by the conservative force is: dWhay, . The work done by the conservative force is equal to decrease in potential energy of the system. Let U be the potential energy and—dU be the decrease in potential energy. Then =dU= dere or, --dU=F.dr . => AU=U,-U,=Sau= fF |GRAVITATIONAL POTENTIAL ENERGY The gravitational force on a ball of mass m is mg, gis regarded as constant near the Earth’s surface (i.e., height ‘h’ of the ball above the Earth's surface is very small as compared to Earth’s radius R(h<< R)) so that variation of gis ignored, Lets take the upward direction as positive, Consider raising a ball of mass m up to a height ‘h’. The work done by the extemal agency against the gravitational. force is mgh, which gets stored as potential energy. Gravitational potential energy as a function of the height f, is denoted by U(h) and itis the negative of work done by the gravitational force in raising the body to that height. U(h) = mghAG Work, Energy and Power Board, JEE (Main) & State Competitive Exams. Ii taken as variable, gravitational force F equals the negative of the derivative of U(h) wart. h. =f The negative sign indicates that the gravitational force is downward. On releasing, ball comes down with increasing speed. If vis the speed just before it hits the ground, then v= 2gh Fi Muttiplying both sides of this equation with >m, we get, 1 qin =mgh Which shows that the gfavitational potential energy of the body at height h, manifests itself as kinetic energy on reaching the ground. Physically, this notion is applicable only to the, class of forces where work done against the force gets’stored up’ as energy. It manifests itself as kinetic energy when external constraints are removed. Mathematically, the potential energy U(2) (for simplicity, in one dimension) is defined if the force F(x) can be written as Fey Oa frac fou U-Ur “The work done by 4 conservative force such as gravity or the spring force depends on the initial and final positions only, ¢.g., an object of mass m is released from rest, from the top of a smooth (frictionless) incined plane of height f, its speed at the bottom = 2gh irrespective of the angle of inclination (itdoes not depend on path taken). ifthe work done or the Kinetic energy depends on other factors such as the velocity or the path taken by the body, the force is known as non-conservative force. Dimensions of energy : [ML’T~] (same for KE, PE and work) 1 unit of potential energy : Joule, the same as kinetic energy or work. Commercial unit of energy is kWh, 1 kWh =3.6 x 10° J To reiterate, the change in potential energy, for a conservative force, AU is equal to the negative of the work done by the force. aUu=-| Fox)dx We saw in the example of falling ball, hw potential energy was converted to kinetic energy. We shall now examine an important principle of conservation in mechanics. ELASTIC POTENTIAL ENERGY OF SPRING “The eneray associated with the state of compression or expansion of a spring is known as elastic potential energy. Consider a spring of natural length /. A block of mass m is connected to he Paring, When the spring is extended from its natural length fy by a distance , it stores a potential energy U;, When the spring is extended from its natural length lo by a distance Xz, it stores a potential energy U2. —,—> “The work done by the spring force on the block, as the extension increases from x4 10 X18 1 . Wey =—3 kE-*1) ‘This work done by spring force is equal to decrease in potential energy stored in the spring. (16038Board, JEE (Main) & State Competitive Exams Work, Energy and Power ia Thus, we have Wey = U, ~ Up Vy, 4 k or, ~yk(xe x1) = Ur Ue 4 eg? or, U2 Ur= Shap ket hx, If the potential energy of the spring is taken as zero, when there is no extension/compression, then _ take x, = 0 and U, = 0 in the above expression to get - A U;-0= dud k or Um? (weer x extension/compression ) 2 from natural length) ht THE WORK-ENERGY THEOREM The work-energy theorem states that the work done by all forces acting on a particle is equal to the change in the particle's kinetic energy. Wea = AKE = nv2?-1 v2 Wei = 3mV?—>mV?- Confining to one dimension, rate of change of kinetic energy with time is = pode) (as mis constant) = Fv (as F=ma= mm (from Newton's II law)) aK dx at Pat . or dK=Fdx wow J ax= f Fax Ba [Integrating from the initial position (x) to the final position (x)] Where K and K, are the initial and final kinetic energies corresponding to x, and x, respectively. ii Fax . . or ]K~K, Thus, [K,-Ki= W]SGD! Work, Eneray and Power Board, JEE (Main) & State Competitive Exams Mlustration 2 Consider a block of mass m placed on a smooth horizontal: Surface [Figure a)}. force Facts on the:block. AtA,.speed ‘ofthe block is ‘uv. When the S int B, is 'V. [Figure (b)] 5 ae Figure (a) Work done by normal reaction : N ‘Work done by component F'sin @ Here, we can see thatthe ofly work done on the blockis by the force F. Let us now find out the change in kinetic energy. iene coloration ofthe block is along horizontal. Let'a’ be the acceleration. So, we have = 248 008 a Also, a=" (As only foree along horizontals F cos 4) ‘Therefore, the above result can be written as LiL mu? = Foos $x S gm ym lace ‘The left hand side is the change in kinetic energy of the body and right hand side is the work done by only force acting along horizontal. n fact, this is also the total work done by all the forces. So, the definition of work done as W= F- S nicely fits into our definition of Work-Energy theorem. Note © InWork-Energy theorem, <2 Geiporate Onfics “Aakash Tower, 8, Pusa Road, New Delt 140005.Board, JEE (Main) & State Competitive Exams Work, Energy and Power Saosin Eras ‘A particle of mass 0.5 kg is subjected to a force which varies with distance as shown in figure Solution : FN, @ 12 x(m) The work done by the force during the displacement from x = 0 to x = 4 and x = 4to x= 12 are respectively (1) 204,400 (2) 404,405 (3) 204,604 (4) 404,805 The increase in kinetic energy of the particle during its displacement from x = 0 to x=12mis () 200 (2) 40 (3) 80d (4) 1203 Ifthe speed of the particle at x = 0 is 4 mvs, its speed at x = 8 mis (1) 8m/s (2) 16 m/s (3) 32m/s (4) 4ms Answer (3) During the displacement x = 0 to x = 4, work done is atx base x height = $ x4%10=204 During the displacement x = 4 to x= 12, work done is W=4™10+ $ x4x10=60J ‘Answer (3) During the displacement x = 0 to x = 12, work done = total area = Pxartoran 1045 4x 10= 80d This is equal to increase in kinetic energy. Answer (2) By work-energy theorem, Wai = AKE During the displacement x = 0 to x= 8 we $x4x 1044 x10=60 J KE= $x05xae= 4) By work-energy theorem, W=KE,— KE; 4 60> 5mv-4 fxosxv=64 =>Vv=16 misAA Work, Energy and Power Board, JEE (Main) & State Competitive Exams A raindrop of mass 2g falling from a height of 1.00 km, hits the ground with a speed of 40.0 ms". (a) _ Find the work done by the gravitational force. (b) Find the work done by the opposing resistive force (g = 10 ms) Solution : (a) Assuming u = 0 and constant g ='10 ms~, the work done by the gravitational force is W, = mgh = 2x 10% x 10x 10° = [20d (b) ak=Emv-0 é = $x2% 10% %40%40 #165 From the work-energy theorem Ak=W,+W, — where W, is the work done by the opposing resistive force. => 1.6=20+W, => [Wa 18.4d The position (x) of a particle of mass 1 kg moving along x-axis at time tis given by (x = $F) metre. Find the work done by force acting on it in time interval from t= 0 to t= 3. Solution : 1 a2" Att=0,v,=0 Att=35, v,=3ms* According to W-E theorem, W=AK =K-K =dmvi-fov?- spe x3?=4.55 POWER Power is defined as the time rate at which work is done or energy is transferred. The average power of a force is defined as the ratio of the work, W to the total time t taken Ww Paap The instantaneous power is defined as the limiting value of the average power as time interval | approaches zero. The work dW done by force F for a displacement df is dW= Fd?Board, JEE (Main) & State Competitive Exams ‘Work, Energy and Power aa | ‘The instantaneous power can also be expressed as. oF P=F- Dak at where 7 is the instantaneous velocity when the force is F. Power is a scalar quantity like work and energy. Its dimensions are [ML?T~] Its SI unit is Watt (W). | Ean Solution : ‘An engine pumps 400 kg of water through height of 10 m in 40 s. Find the power of the ‘engine if its efficiency is 80% (g = 10 ms”). Work done by engine against gravity W. = mgh = 400 x 10 x 10= 40 kd 40x 10° Power used by engine = @ = 40% 10 wesw At 40 If power of the engine is P, then 80 _ 4000x100, 100, Px agg 71 KW = P= W or kW = [125K A particle is moving along x-axis under the action of a force, F which varies with its position Solution : (3) a8 Fe. Find the variation of power due to this force with x. Ik Fax > acx => a= (where kis a proportionality constant) dv xv ny a dtd ™ wie ox. > a => vdv = kody = Suave khan Po x'®)SEUNG Work, Energy and Power Board, JEE (Main) & State Competitive Exams ‘An elevator that can carry a maximum load of 1500 kg (elevator + passengers) is moving | up with @ constant speed of 2 ms". The frictional force opposing the motion is 3000 N. ” Find the minimum power delivered by the motor to the elevator in watts as well as in horse power (g = 10 ms) q Solution: The downward force on the elevator is ‘ F=mg+F, = (1500 x 10) + 3000 = 18000 N The motor must supply enough power to balance this force. Hence, P=F.7= 180002 =[s6000W] =—7-- =[48.3hp [Work, Positive and negative work, Notion of work and kinetic energy + The work-energy theorem, Kinetic energy, Work done by a variable force] aa A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If work done is 25 J, then the angle which the force makes with direction of motion of the body is 4 (1) 60° (2) 90° 3) 0° (4) 30° 3 A force (mv) is acting on a body of mass m moving with a velocity vin a circle of radius r. What ig the work done by the force in moving the body over half the circumference of the circle? (1) “xree Board, JEE (Main) & State Competitive Exams Work, Energy and Power ae XC Awoman lifts a barbell 2.0 m in 5.0's. Ifshe lifts it by the same distance in 10 s, the work done by her is 4) Four times as great (2) Two times as great () (3) The same (4) Half as great XC Aparticle is moved from (0, 0) to (a, a) under a force F = (3i + 4)) from two paths. Path 4 is OP and path 2 is OQP. Let W, and W, be the work done by this force in these two paths. Then (1) Maw, pea) (2) W, = 2W, xd se (3) 2W, = W, x (4) W, = 4, ® t, Ss a. Aforce that object A exerts on object B is observed over a 10 second interval, as shown on the graph. How much work did object A do during that 10 s? (1) Zero (2) 1255 (3) 250 (4) 505 A particle moves along the x-axis from x = x, to x =x, under the influence of a force given by F'= 2x. Work done in the process is (1) Zero (2) x2 — x? (8) 2x2 (x2 x1) 4) 2x1 04— 7. _Abody of mass 5 kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on itin the direction of motion of the body for 10 s. The increase in its kinetic energy is (1) 284 (2) 32d (3) 38d (4) 444 4 kg mass moving with speed 2 mis, and a 2 kg mass moving with a speed of 4 m/s, are sliding over a horizontal frictionless surface. Both objects encounter the same horizontal force, which directly opposes their motion, and are brought to rest by it. Which statement best describes their respective stopping distances? : (1) The 4 kg mass travels twice as far as the 2 kg mass before stopping (2) The 2 kg mass travels twice as far as the 4 kg mass before stopping (3) Both masses travel the same distance before stopping (4) The 2 kg mass travels * qeater than twice as far as the 4 kg mass X= 3t- 4 + , where x is in metre and fis in sec. The work done during first 4 s is (1) 570.mJ (2) 450 mJ (3) 490 mJ (4) 528 mJ 10. Awater pump is used to deliver the water at constant rate. By what factor power of pump should be increased to increase the rates.n times? : (jn (2) r? (3) (4) ne 7AE! Work, Energy and Power Board, JEE (Main) & State Competitive Exams. 37-12], undergoes a displacement of d=4/. If its kinetic energy 4% A particle which is experiencing a force, given by the particle had a kinetic énergy of 3 J at the beginning of the displacement, what is at the end of the displacement? (1) 155 (9d (@) 124 (4) 10J 15 A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (lake g = 9.8 ms (1) 4627.20 (2) -62720.0) (3) 627.25 (4) 784.0 J Work Done by Some Common Forces In Newton's laws of motion we have studied about different types of force like tension, normal reaction, friction force etc. Now let us find out the work done by these different types of forces. Work Done by Tension Two blocks having masses M and m are connected by a light inextensible string and the system is pulled by a horizontal force F on a smooth horizontal surface as shown. Find work done by the tension an two blocks as the system moves by a distance d. alrebeeakath Solution: (i) Work-done by the tension on the block of mass m AW, = Td. : (i) Work-done by the tension on the block of mass M AW,, = Tdoos180° = -Td (iii) Net work done by the string AW= AW, + AWy =Td+[-Ta}=0 7 Bob of a simple pendulum is released from the shown position A. Find work done by the string on the bob as it reaches the point B. Solution: __ In this case, For every small displacement of the bob dx, tension of the string s perpendicular to the displacement therefore for every small displacement work done by the string dW/= Tdxcos5 = 0 Therefore total work done by the tension in the process AW=0 . C A block of mass 'm’ is suspended from the ceiling of a lift by a massless rope. The lift is going down with an acceleration g Find the work done by various forces acting on the block as it goes down through a height h. T (by rope) Solution: The FBD of the block is shown. By Newton's 2” law, ae |e =f ing T= mae 2 T : =>T= 2 hl Work done by the various forces : (1) Wg = mg * h cos 0° = mgh yw, Jmgh mig (weight) x hx cos 180° iBoard, JEE (Main) & State Competitive Exams Work, Energy and Power |igisaaane Work Done by Normal Reaction Two blocks having masses M and mare placed on a smooth horizontal floor and a horizontal force Fis applied on the system as shown. Find work done by the normal reaction between the blocks on two blocks as the system moves by a distance d. = (i) Work done by the normal reaction on the block of mass m AW, = Nd N (ii) Work done by normal reaction on the block of mass M AW, = Ndcos180° = -Nd d (iii) Net work done by normal reaction between two blocks on the system of two blocks, AWyqe = AWp, + AWe, = Nd + (-Nd) = 0 ‘A block of mass M moves on a smooth horizontal surface by a distance d under the action Of force F as shown. Find work done on the block by the normal reaction inthe process. - d Solution: Angle between normal reaction and displacement is 90° therefore work done by the normal reaction. qd AW = Ndcosg0° = 0 Work Done by Friction (a) Work done by static friction Abicck of mass Mis placed on a rough horizontal platform. M. sole ce (i) A.pulling force horizontally as shown. If F is less than or equal to maximum possible frictional force between block and floor (limiting friction) then the block will not slip and work done by frictional force acting on the block will be zero as displacement of the block will be zero. : F F - OF, Xena = ye Figure shows a block of mass 'm' resting on a smooth horizontal surface. Itis connected to a rigid wall by a massless spring of stiffness ‘k’. The spring is in its natural length. A constant horizontal force F starts acting on the block towards right. Find (i) speed of the block as it moves through a distance x, (i) speed when the block is in equilibrium and (ii) maximum extension produced in the spring. Solution: When the block has moved through ‘x, its speed is (say). ‘The FBD of the block in this situation is shown. . () By work-energy principle, W= AK 7 Wet Wop: ym -0 os = Fax cos0r f oxa= tm? |EBB Work, Energy and Power . Board, JEE (Main) & State Competitive Exams (i) At equilibrium, net force is zero. FE Fe kxor x= QF x= Now, v= m 2A ved kk WE mink (ii) Atthe maximum extension, speed of the block becomes zero. v=0 2Fx- hoe 2F or, = 0 = x= 5 m kK Work Done by Pseudo Force When we observe the displacement of a body under the action of a force from a non-inertial frame then to calculate total work done we have to consider work done by pseudo force also. A block of mass M = 5 kg is placed on a smooth horizontal surface and a force F = 25 N starts acting on it parallel to the smooth plane. Find net work done on the block as seen by an observer accelerating in the direction of the force with acceleration 2 m/s? as the block gets displaced by 5 m in the direction of the force with respect to the observer. M FE Smooth K_. a= 2s Observer Solution: F.B.D. of the block wirt. the observer. Work done by weight of the body-and normal reaction will be zero. Now, net work done AW = AW. + AWs, = Fd-Mad (25 x 5) — (5x 2x 5)» Board, JEE (Main) & State Competitive Exams Work, Energy and Power Feil ‘done by tension, Work done by friction, Work done by spring force, Work done by pseudo force] 13. Aboxis dragged across a floor by a rope which makes an angle of 45° with the horizontal. The tension in the rope is 100 N, while the box is dragged 10 m. The work done is Mt) 607.1 J (2) 707.15 (3) 1414.25 (4), 9004 14.. Ablock of mass mis suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration ‘a’. Work done during t s by the tension in the thread is () Forayae @) Z-aae (8) Foatt (4) Zero 180 Abody of mass itis dropped from a height'n. The speed with which body hits the grounds {0.9 gh. Find the work done by air-resistance during its fall. (1) 40.55mgh (2) -0.55mgh (3) +0.45mgh (4) -0.45mgh 16, The 300 N crate slides down the inclined curved path from A to Bin the vertical plane as shown in the figure. if the crate has a velocity of 1.2 m/s down’the inclined at point A and 8 m/s at B. Find the work done against friction during the motion. (1) 860 N m (2) 987 Nm (3) 1200 Nm oi ~ (4) 500 Nm \__,., “ 17. Acar moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. Ifthe same car is moving with a speed of 80 knv/h, what is the minimum stopping distance? (1) 8m (2) 6m (3) 4m (4) 2m 18. A block is attached to the end of an ideal spring and moved from coordinate x; to coordinate x,. The relaxed position is at x = 0. The work done by spring is positive if (1) x= 2 om and x,= 4 om (2) x (3) x=-4 0m and x,=-20m (4) 4= 2m and x/= 4 cm om and x,=—2 cm 187 A spring of natural langth L compressed to length 7L/8 exerts a force Fy. The work done by the spring in restoring itself to natural length is (1) FyLl2s (2) FoLt6 (3) 3F L256 (4) Foll8 20. ‘An ideal spring is used to fire @ 2 g block horizontally across a frictionless table top. The spring has a spring constant of 80 N/m and is initially compressed by 3 cm. The speed of the block as it leaves the spring is (1) Zero (2) 2* 107 mis (3) 6 * 10? mils (4) 36 mis (5) 6 mis 21. A 2g block attached to an ideal spring with a spring constant of 80 N/m oscillates on-a horizontal frictionless surface. When the spring is 4.0 cm shorter than its equilibrium length, the speed of the block is 417 mis. The greatest speed of the block is. (1) 4ms (2) 6 ms (3) 5 mis (4) 9 ms (5) 3 mis22. 23. BG} Work, Energy and Power Board, JEE (Main) & State Competitive Exams Two blocks A and B of masses rm and 2m respectively are connected at two ends of an ideal spring of spring constant kas shown in figure. Block A is pressed down by a force F and now itis in the state of ‘equilibrium. When the force Fis suddenly removed, block B jumps-off from the ground. The minimum ‘compression in the spring at the given equilibrium position is E 3mg 4mg Oe Oe « mg 2mg @ = a= Ablock of mass m is kept on a platform which starts from rest with constant acceleration 2 upward, as shown in fig. Work done by normal reaction on block in time tis mgr? Pe @ a 5 fers @o we a 72 THE CONSERVATION OF MECHANICAL ENERGY For simplicity, we are considering one-dimensional motion. Suppose that a body undergoes displacement Ax under the action of a conservative force F. From the W-E theorem, ak= { F(ndx If the force is conservative, the potential energy function U(x) can be defined such that AU =~ f F(x)ax Adding the above two equations, we get AK+AU=0 or A(K+U)=0 = K+U, the sum of the kinetic and potential energies of the body is a constant. ‘Over the whole path x; to x; K+ Ux) = Ky + Ux) ° I ‘The quantity K + U(x) is the total mechanical energy of the system. Kinetic energy K and the potential energy U(x) may change individually from point to point, but the sum is a constant. “The total mechanical energy of a system is conserved if the forces doing work on it, are conservative.” Let us consider a ball of mass m being dropped from a cliff of height H. The conversion of PE to KE for a ball of mass m dropped from height H. Total mechanical energies &;, E, and E, at the indicated heights zero (ground level), h and H, are E,=mgH (PE at height H as KE = 0) E,= mgh+Smv3 (Sum of PE and KE, PE, = mgh, KE» => mvi ) mv; (KE at the ground as PE = 0) sprerenstililit sosccou cesonasneomnannseet:Board, JEE (Main) & State Competitive Exams Work, Energy and Power Se where v, and v,are velocities at height h and at the ground. The constant force is a special case of a spatially dependent force F(x). Hence, the mechanical energy is conserved thus, E,= E, or = mgH= dmv v= fegH - Aresult that we got earlier for a freely falling body. Further, 6, = 6, = mgH=mgh+Lmvi or R= 2g(H-h) A familiar result from kinematics. At height H, the energy is purely potential (KE = 0). At height A, the energy is partially converted to kinetic. At ground level the energy is fully kinetic (PE=0). * This illustrates the conservation of mechanical energy. Abody of massWork, Energy and Power Board, JEE (Main) & State Competitive Exams ‘particle is set moving with kinetic energy K straight up a rough inclined plane, ofinelination «and cosficient of ition y. Prove that the work done against friction before the particle first comes to rest is Po — sina + pcos Let the mass of the particle be m, the normal reaction on the plane be R, the initial velocit be v, and the distance travelled up the plane before it first comes to rest be s. By energy conservation, Solution : 1 K= mv? = mgssina + uRs tly at rest Where, R= mg cos a Hence, K=mgs(sin a + 4 cos «) The work done against friction is wR = umgscosa= coe: Figure shows two discs of same mass m. They are rigidly attached to a spring of stiffn k. The system is in equilibrium. From this equilibrium position, the upper disc is pres: down slowly by a distance x and released. Find the minimum value of x, ifthe lower disc Just lifted off the ground. Kucoso Let us first find out the minimum elongation (x’) needed to lift the lower disc from the grount Solution : The free body diagram of the lower disc at this instant will be z t . => ke = mg 7 + mg. ed TI aaaet N=0 mg (lower disc) “The system given in the question isin equilibrium thus forthe equilibrium of upper disc ni ‘compression (xo) in spring is = y= mg ak kx Now from this equilibrium state spring is further compressed by.an amount x and releas' Now let us apply energy conservation from this compressed state and final state. Let us take the reference for gravitation potential energy at unstretched position then 1 ka xa)? malo) = iat mgx’Board, JEE (Main) & State Competitive Exams Work, Energy and Power Cae pag) Now, x= mg xo % On solving, we get _2mg xk Energy Diagrams spring is fixed at one end and the other end is attached to mass m which moves back and forth with an amplitude A (i) Elastic potential energy U0) = fhe? If frictional force is neglected, we have the energy of the system being conserved, E = constant When x= 0, U =Eand k= 4 me, vis the greatest velocity in the whole process of oscillation. Atx=-A, E= max (U), K=0=>v=0 au “ax = The negative slope of the curve U(x) against x (i) F du (ii) At the origin [= 0, itis the equilibrium position, The resultant force is zero. (iv) When x > 0,22 me > Ovhence F<. This represents that the restoring fice is pointing to the left. al (v) When x< 0 <0, hence F > 0. This represents that the restoring force is pointing to the right. - (vi), When x= 0, itis called equilibrium position.a) Work, Energy and Power Board, JEE (Main) & State Competitive Exams The given curve shows a profile of an arbitrary potential energy with the posit * Conservation of mechanical energy E= K+ U(x) = constant 1 k= 5mv20 (i) (f &=&;initially, then the particle will stay at point xo, otherwise K has to be negative. (i) When E= E,, the motion of the particle is confined between x, < x < x2, x; and x, are called turning points. . (ii) = E,, depending on the initial condition of x, the particle wll ether move in region | or region Il. (iv) E=E,, the particle will move between x = x3 and «. (v) Unstable equilibrium position - Any maximum in a potential energy curve is an unstable equilibrium position. The graph shows the potential energy of a system of two ions as a function of their interionic separation. Note that: 1 aJ= 1% 10~® Jand 1nm=1 «10° m. PEfad - 10: ott t of { | Tt | <2 ay! 0.0 Ot 0.2 03 04 05 06 07 Separation/inm The system of the two ions has a total mechanical energy of -0.5 x 10-7 J. What is the range of possible values for the distance between the two ions? At what separation do the ions have minimum kinetic energy? What is the maximum possible kinetic energy of the 4 ions? What energy is required to break apart this ionic molecule? Solution: (i) 1.7 x 10- m to 5.0 x 10 m (Cutside this range the potential energy would need to be greater than the total energy, that is the kinetic energy would have to be negative which is an impossible situation). 4.7 * 10° m or 5.0 x 10-? m (K.E. = 0) 1.5% 10-7 J (at d= 3.0 x 10° m) 0.5 x 10-"” J. (The total mechanical energy must be increased to 0)Board, JEE (Main) & State Competitive Exams Work, Energy and Power Saas [Conservation of mechanical energy, General study of the potential energy curve] 24. Asimple pendulum consists of a 2.0 kg mass attached to a string. Itis released from rest at X as shown. Its speed at the lowest point Yis (1) 0.90 mis (2) 86 mis (3) 3.6 mis (4) 6.0 mis 7 (8) 36 mis 25. A block is released from rest at point P and slides along the frictionless track shown. At point Q, its speed is . (1) 2aVhi=Fe (2) 2g(hy~ ha) hy - fe | (a) ahd) . | Z 2g Ground level (4) f2g(h.=he) 26. Aballis held at a height H above a floor. It is then released and falls to the floor. If air resistance can be ignored, which of the five graphs below correctly gives the mechanical energy E of the Earth-ball system as a function of the altitude y of the ball? PI y elH - e| eI | &| q 1! y TY wy rma ' iM \V v at (2) @) im @) Iv 6) Vv 27. For a block of mass m to slide without friction up the rise of height h shown, it must have a minimum. initial speed of _[w}r 7h @) 29h ; ) eal ) (4) 242gh (©) 29hBe Work, Energy and Power Board, JEE (Main) & State Competitive Exams 28. A2.2 kg block starts from rest on a rough inclined plane that makes an angle of 25° with the horizontal. The coefficient of kinetic friction is 0.25. As the block goes 2.0 m down the plane, the mechanical energy of the Earth-block system changes by (1) Zero (2) -9.85 (3) 985 (4) -18 (6) 18d 29. The potential energy of a particle moving along the x-axis is given by U(x) = 8x? + 2x", where Us in joules and xis in meters. If the total mechanical energy is 9.0 J, the limits of motion are (1) 0.96 m, +0.96m (2) -2.2m, 422m — (3) -1.6 m, 41.6 m (4) 0.96 m, +2.2m (8) -0.96 m,#1.6 m 30. The graphs below show the force acting on a particle moves it along the positive x-axis from the origin to x= x;. The force is parallel to the x-axis and is conservative. The maximum magnitude F, has the same value for all graphs. Rank the situations according to the change in the potential eneray associated with the force, least (or most negative) to greatest (or most positive) A FJ xx xX 1 2 (171,23 (2) 1,32 8) 2.3.4 (4) 3,24 (5) 2,1,3 31, The first graph shows the potential energy U(x) for a particle moving on the x-axis. Which of the following five graphs correctly gives the force F exerted on the particle? UI Paraboia x F F F F F x x x x x W i Vv v at 2) (3) @V . Vv 32. The diagram shows a plot of the potential energy as a function of x for a particle moving along the x-axis. The points of stable equilibrium are (1) Only a (2) Only b (3) Only ¢ (4) Only @ Y if (6) band d abo. de 33. A particle is released from rest at the point x = a and moves along the x-axis subject to the potential energy function U(x) shown. The particle (1) Oscillates back and forth between x = a and x= (2) Oscillates back and forth between x= @ and x= 0 (3) Moves to infinity at varying speed (4) Moves to x = e where it remains at rest (6) Moves to x = b where it remains at rest xBoard, JEE (Main) & State Competitive Exams Work, Energy and Power eee MOTION IN A VERTICAL CIRCLE Leta particle P be suspended in a vertical plane, by a massless, inextensible string froma fixed point . In Figure the string is vertical with P vertically below the point of suspension 0. @ () The particle is in equilibrium: Let the particle be given an initial velocity V7, in horizontal direction, as shown in figure (b). The particle moves along a vertical circle (radius = length of the string). The point of suspension is the centre of the circle. This motion has to be non-uniform circular motion. Velocity of the particle changes both in magnitude and direction. Speed of particle decreases continuously as it moves up the circle (i.e, from P—> Q — R) due to the work done against the force of gravity speed of particle increases continuously as it moves down the circle (Le. from R-> @ > P) due to work done by the force of gravity on the particle mgsind. mg Fig. : (c) To get the basic characteristics of vertical circular motion, consider an instantaneous position of a article, say at L. This is shown in Fig. (c), where the string makes an angle 0 with the vertical line OP. At this position, the forces acting on the particle are (Weight (= mg) acting vertically downwards (i) Tension (= 7) acting along LO (in the string) Direction of the instantaneous velocity v is along the tangent to the circle at L. The corresponding instantaneous centripetal force = ~ twhere r(= length of string /) is radius of the circular path.] acts along LO. r Taking components of mg, mg cos® acts opposite to LO and mgsin0 acts opposite to 7 (along tangent to the circular path at L, ie., perpendicular to LO) The net force towards centre ofthe circle (along LO) = T- mgcos0 this is necessary centripetal force -*) or mv So, “—=T-mgcose T=2 +mgcosd 4) Taking horizontal direction at the lowest point P, as the position of zero gravitational potential energy, as per the law of conservation of energy Total energy at P = Total energy at L 1 oat mui +0=>mv'+mgh (2)‘Work, Energy and Power Board, JEE (Main) & State Competitive Exams Where MP = h, is the vertical height of the particle above P. OM = OL.cos® = rcosé (from right angled AOML) MP = h= OP - OM =r- reos@ = 1(1 - cos®) or h=1(1~cos0) (8) From equations (2) & (3) sawi = A av?+mgr(' cos 0) 2 2 or vi=Vv'+2gr(1—cos®) wf) Put the value of v from equation (4) in equation (1) m 5 [vi-2gr(1 - cos®)] + mgcos® ma -2mg(1-cos0) +mgcos® 2 =" —2mg+3mgeose AB) __ This equation gives tension T as a function of 8. We can use it to see the details when the particle is at (|) lowest (P) (ii) mid way (horizontal) (Q) and (ii) highest position (R) ALP, 0=0 The tension in the string = Tp mvp To= “tt —-2mg+3mgcos0" rom equation (5) F mi Te= th smg (8) {At Q, 6 = 90° string isin horizontal position (momentarily). Fi Let 7» be the instantaneous velocity at Q and let To be the instantaneous tension in the string. tg Using equation (5), mv Tae et amg +3 mgoos 90° Ta= et -2mg fT) ‘The change in the tension, as the particle moves from PtoQ 2 2 = Tp-To= (172 ma) -(™ -2ma) Using equations (6) and (7) Tp-To=3img ‘At highest point R, @ = 180° My RBoard, JEE (Main) & State Competitive Exams Work, Energy and Power Sissies 9 Let 7; be the instantaneous velocity at R and let Tx the instantaneous tension in the string. Using equation (5), 2 mv 7 -2g + 3mgcos 180" (8) The change in tension in the string as the particle moves from P'toR =Tp-Tr mvt mi = (7 +mg}-(7%—smg) Using equations (6) & (8) Tp-Ta=6mg => The tension in the string is maximum at lowest point P. And The tension in the string is minimum at highest point R. This is so because at the highest point, a part of centripetal force, needed to keep the particle moving in circular path is provided by weight (mg) of the particle. From equation (8), we find that T, can be (i) positive (i) negative or (i) zero, depending on the value of v,..When T, becomes negative, string slackens and the particle will fall down before completing its circular path. => Minimum value of Tz should be zero for completing the vertical circle vin | So, (Tr)min = -5mg=0 from equation (8) | (Wa)nin = 459F (9) | Using equation (4), the minimum speed, which the particle must have at the highest point R, so that it completes the vertical circle, is given by (V1)rin = (Vs)min + 2gr(1 — cos 180°) (as 6 = 180° at R) 5gr= (Vs) 49r Using (9) oF (Wann = 19r When the particle completes its motion along the vertical circle, it is called ‘looping the loop’. For this, minimum speed at the lowest point must be {Sgr . Let us calculate tension in the string, when the particle is just able to do ‘looping the loop’ corresponding to Y= (Vs)min = 4597 From equation (6) m ~Te= 2 (6gn) +mg=6 mg From equation (7) m Ta=— (gr) 5mg=0SSE: Work, Energy and Power _ Board, JEE (Main) & State Competitive Exams [EQEEETE) 4 particle of mass m slides down a smooth curved surface, which ends into a vertical loop Solution : EEE Asmall stone of mass 0.4 kg tied to a massless inextensible string is made to loop the loop. ~ Speed at highest point = {gr = {104 tadius R. What should be the minimum height h such that if the particle released from it, does not fall at the uppermost point of the loop? R Ifthe particle does nét fl atthe highest point then normal reaction at highest point A N20 7 F.B.D. at topmost point A mv R A vwarct. ground . | N79 = = A+ mg= v= vag, when Nq=0 Vein = QR wal) Applying conservation of energy between initial and final point. K+U,=K,+U, [Assuming final point as a reference] 0+mg(h-2R)= tm? 2g(h-2R) = oR 2h-4R=R uouUUY Radius of the path is 4 m. Find its speed at the highest point. How would this speed change ‘if mass of the stone is decreased by 10%? (g = 10 ms). 6.32ms"}. Itdoes not depend on the mass ofthe stone. So, ths speed remains the same on changing mass of the stone.Board, JEE (Main) & State Compe' Work, Energy and Power (gauss The figure shows a smalll ball of mass 0.1 kg placed on a smooth plane surface OA which acquires a semi-circular shape ABC of radius 2m. The ball just touches a light spring of stiffness 1000 N/m. The ball is pushed to the left so as to compress the spring by a distance and released. This ball then starts moving towards the circular track ABC. (g = 10 m/s?) c ‘Smooth A (a) Find the minimum work done by external agent to push thé ball to the left through 50cm. (b) Ifthe ball is pushed to the left by 5 cm and reléased, calculate (). Normal force on the ball just after crossing A. (i) Maxinum angle covered with respect to PA on the circular track before it comes to rest. () What is the minimum distance Xp, by which the ball should be pushed to the left and released so that it can reach up to C? (@) If the ball is pushed to left by 0.7 Xmiy calculate () Reaction force between ball and track at point B (i), Maximum height attained by the ball above horizontal stirface OA. Solution: (a) When the work done by external force to push the ball against the spring is minimum, there must be no kinetic energy associated with the ball. The work done is solely responsible forthe potential energy. Wash? x 1000 «(y= 1.25) (0) As the ball leaves the spring, it will be moving towards right with a speed v, such that the potential énergy of the spring is converted to kinetic energy of the ball. 1 1 > qi = he [kK = ven As, k= 1000 Nim, m= 0.1 kg, x= 5 om () As it crosses A, its path becomes circular and it experiences a centripetal acceleration a, =~ fn By Newton's 2” law, N-mg=ma, me 25 N= mgt = 0.1 x 1040.1 x =141,25=225N ed . (i) As the ball rises on the track, its gravitational potential energy increases and kinetic energy decreases. As it comes to rest, kinetic energy becomes zero. Let ithappen at a height h.2BE Work, Energy and Power Board, JEE (Main) & State Competitive Exams () @ By conservation of energy, c mgh+0= +L ¥v_ 8 2 2g “2x10 20° '25™ ° arf B Now, cogo-=2zf.= 271.25 _ 0.75 ° 2 20 = 0-008" (3) A Let the spring is compressed by Xin and the speed of the ball after leaving contact t oto giv => kan h with spring is v, then by energy conservation f= v= Xmin Now, to complete circular motion minimum speed at A should be equal to 15gR.. = san = f5GR m => Xp = 0.1 ‘When the spring is compressed by 0.7 Xm = 0.07 m, the speed acquired by the ball is v= xf 0.07 x 5 (As the ball reaches B, its speed becomes v, such that 5 gm Ling mgR (by conservation of mechanical energy) — g. Va=4v?—2gR = {49-40 aN ve= 3 mis The FBD of ball at Bis shown. The radial acceleration at this instant is a, = “ =3 mis? N By Newton's 2° law, mg, 2 wae — N="BP=0.1x5=045N ii (ii) As the speed at lowest point is less than what is required to complete the circular track and reach C, the ball will leave the track before it reaches C. Let it occur at D such that OD makes an angle @ with OC. Let vp be the speed at this position. The height at Dis h = R(1 + cos0) By conservation of mechanical energy, 1 1 qin’ => mvp+mgR(1 +c0s8) ‘At D, FBD of the ball is shown. (1)Board, JEE (Main) & State Competitive Exams Work, Energy and Power (SQ ussananarsy As ball leaves the track at D, normal reaction is zero. By Newton's 2 law in radial direction, 2 mvp R vg = gRcos (2) From (1) & (2), ime = J marcose+maR(1 +c0s@) mgcos6 = => ¥—2gR = 3gRcos0 W-29R _7?-2x 10x 3gR Bx 10X2 =R(14+3)- 3B x20 és h=R(1+3) 30 *2=23m From point D, the ball moves in a parabolic path under the action of gravity alone. From this point onwards, it rises further to a height H given by, => cos¢= _vasin?o 2g gRcosesin?o SH [from (1)] (1h) = Best 118 400)” 20 “400 ~ 8000" 1173 ‘8000 Maximum height above OA is 2.3 + =2.45m EXERCISE 5.4] [Motion in a vertical circle, Elastic potential energy of spring] 34, Aball of mass m, at one end of a string of length L, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. The speed of the ball at the bottom of the circle is im (1) ok (2) Bat (3) fagh (4) Bot ©) fot 35. Ablock of mass m slides down on a smooth parabolic surface from height has shown. Find its speed at lowermost position (1) {gh (2) sgh . (3) ah (4) Data insufficientPAG: Work, Energy and Power Board, JEE (Main) & State Competitive Exams 36. 37. 39, 40. At. 42. 43, ‘The string in the figure is 50 cm long. When the ball is released from rest, it will swing along the dotted arc. How fast, in m/s, will it be going at the lowest point in its swing? (1) 2.0 (Q) 22 , Bt @) 3.4 ne - (4) 44 Aparticle of mass mis fixed to one end of a light rigid rod of length ¢ and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be (1) Zero (2) (gt (3) A5gt (4) 2gt ‘Assmall object of mass m, at the end of a light cord, is held horizontally at a distance r from a fixed support as shown. The object is then released. What is the tension in the cord when the object is at the lowest point of its swing? oe (2) mg \ " @) 2mg (4) 3mg When a block of mass ‘m’is released from point ‘4’ on the surface of a smooth bow, the normal reaction force on it at B would be equal to 7a . A (1) mg (2) 2mg R (8) 38mg (4) Zero a ‘Ablock of mass m is released from a height'R on the frictionless incline as shown. The incline leads to a circle of radius R/2. The maximum height attained by the block is (1) Less than R/2 (2) Greater than R An (3) Equal to R (4) Less than R but greater than R/2 Assmall ball of mass m is released from rest from top of smooth track which has circular part at other ‘end as shown in figure. The normal force exerted by track on the ball when it reaches to point A, is (1) mg mg Q > (3) 2mg ~ (4) Zero In the spring mass system shown in the figure, the spring is compressed by x5 = mg/3k from its natural length and the block is released from rest. The speed of the block , when it passes through the point | P at means positiori, is, A km enter] ony iB 21 Hm EI @ ofr @) ofa @ ose ) oe Abody’is thrown vertically upward from the surface of earth with 10 J.of kinetic energy. The potential "| energy (taking reference level at point of projection) at the highest point of its journey is (1) 9d (2) 105 (3) - 100 @) -8J I } a