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GR #2 (NLM, Friction)

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0% found this document useful (0 votes)
336 views5 pages

GR #2 (NLM, Friction)

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sssameera2121
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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GUIDED REVISION ENTHUSIAST & LEADER

Target : JEE(Main + Advanced) - 2025 COURSE

PHYSICS GR # 02 NLM, FRICTION AND CIRCULAR MOTION

1. The system of two weights with masses m1 and m2 are connected with weightless spring as shown. The
system is resting on the support S. The support S is quickly removed. The accelerations of each of the
weights right after the support S is removed are.

(m1  m 2 )g (m1  m 2 )g
(A) a1 = 0, a2 = (B) a1 = 0, a2 = m1
m2

(m1  m 2 )g
(C) a1 = m1 , a2 = 0 (D) a1 = 0, a2 = 0

2. Consider a pendulum, consisting of a massless string with a mass on the end. The mass is held with the
string horizontal, and then released. The mass swings down, and then on its way back up, the string is
cut at point P when it makes an angle of  with the vertical. What should  be, so that the mass travels the
largest horizontal distance from P by the time it returns to the height it had when the string was cut?

 cut

 1  1  1 
(A) tan 1 3 (B) tan 1   (C) tan   (D) tan–1(1)
 3  2
3. A man M slides down a curved frictionless track, starting from rest. The curve obeys the equation
x2
y= . The tangential acceleration of man is :-
2

gx g gx
(A) g (B) (C) (D)
x2  4 2 x2  1
4. A ball of mass m is placed on a sledge of mass M. The sledge moves on an inclined plane A of angle of
inclination . Inclined plane A is moving with constant horizontal acceleration ‘a’ towards left (Neglect
a
friction everywhere)

(A) Normal reaction force between m & M is m(gcos + asin) m


(B) Normal reaction force between A & M is (M+m) (gcos + asin) M
A
(C) aM/A = am/A = (gsin – acos) 
(D) aM = am = sin  g  a2 2

PHYSICS/GR E-1/5
GUIDED REVISION
ENTHUSIAST & LEADER
Target : JEE(Main + Advanced) - 2025 COURSE
5. Two blocks m1 and m2 are connected with a compressed spring and placed on a smooth horizontal
surface as shown in figure. Force constant of spring is k. Under the influence of forces F1 and F2. At an
instant blocks move with common acceleration a0. At that instant force F2 is suddenly withdrawn. Mark
CORRECT option.
a0
F F1 F2
(A) Instantaneous acceleration of m1 is a0 – 1 m1 m2
m1

F2
(B) Instantaneous acceleration of m2 is a2 = a0 +
m2
(C) Instantaneous acceleration of m1 is a1 = a0
(D) Spring force is Fspring = m2a0 + F2
6. Two monkeys (1) & (2) of same mass m = 1kg are hanging on the strings such that block of mass 2 kg
remains at rest and it is given that monkey (2) is just holding the string. Then which of the following
statement(s) are CORRECT (g = 10 m/s2):-

(1)
m
(2)
2kg m

(A) Acceleration of monkey (2) is 10 m/s2 upwards.


(B) Acceleration of monkey (1) is 30 m/s2 upwards.
(C) Acceleration of monkey (1) with respect to his rope is 35 m/s2 upwards.
(D) Acceleration of monkey (2) with respect to his rope is 50 m/s2 upwards.
7. A horizontal force F is applied on a ring of mass m1 constrained to move on a horizontal smooth wire.
The hanging mass m2 is connected to ring with a massless rod and it maintains a constant angle with
vertical. Mark CORRECT option :-
(A) Force on m1 by wire is less than (m1 + m2)g m F 1

m 2F 
(B) Net force on m2 is m  m
1 2
m2
(C) Tension in rod is m2g sec 
(D) F = (m1 + m2) g tan 
Paragraph for question no. 8 and 9
Friction is a force that aids us daily, in fact so much so that we don't even pause to appreciate its
importance. We would not be able to wear pants or jeans without friction. We would have to live like
naked in the jungle. The shirt is supported at our shoulders. But if we stand up the weight of our jeans is
to be supported by a vertical force. The surface of our waist and thighs can be approximated to be like
curved surface of a vertical cylinder. The force of friction acting between this curved surface and jeans
balances the weight of the jeans. To understand this mathematically, let us consider a vertical man
whose waist is a rigid cylinder having a circumference of 90 cm. He wears a jeans of mass 500 gm using
an elastic massless belt which can be assumed to be on elastic string of force constant 500N/m and
circumference 85cm when not extended. The coefficient of friction between waist and jeans is 0.5.

E-2/5 PHYSICS/GR
GUIDED REVISION ENTHUSIAST & LEADER
Target : JEE(Main + Advanced) - 2025 COURSE
8. What is the friction force acting between the man's waist and jeans?
(A) 5N (B) 12.5 N (C) 7.5 N (D) 25 N
9. When the man stands in an elevator going up with a high acceleration, his jeans start sliding down. What
can be the minimum acceleration of the elevator?
(A) 147 m/s2 (B) 288 m/s2 (C) 87 m/s2 (D) 24 m/s2
Paragraph for Question Nos. 10 & 11
A block of mass 1 kg is placed on a rough horizontal surface. A spring is attached to the block whose
other end is joined to a rigid wall, as shown in the figure. A horizontal force is applied on the block so
that it remains at rest while the spring is elongated by x. Let Fmax and Fmin be the maximum and minimum

 mg 
values of force F for which the block remains in equilibrium. For a particular x  where x   ;
 k 
Fmax  Fmin  2 N . Also shown is the variation of Fmax  Fmin versus x, the elongation of the spring.

Fmax + Fmin

5N
K
1kg F
x
0.1m
10. The coefficient of friction between the block and the horizontal surface is :
(A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4
11. The value of Fmin , if x = 3 cm is :
(A) 0 (B) 0.2 N (C) 5N (D) 1N
Paragraph for Questions 12 to 14
In the figure, a horizontal force of 100N is to be applied to a 10kg slab that is initially stationary on a
frictionless surface. A 10kg block lies on the top of the slab, there is no information about friction and
coefficient of friction between the block and the slab.
Block

Slab 100N

12. What can not be a possible value of the acceleration of the slab ?
(A) 7 m/s2 (B) 10 m/s2 (C) 2 m/s2 (D) 9 m/s2
13. What can not be a possible value of the acceleration of the block ?
(A) 4 m/s2 (B) 3 m/s2 (C) 10 m/s2 (D) zero
14. If the ground and the top surface of the slab both are rough, which of the following can not be a possible
free body diagram ? f is friction between block and slab, N is normal between block and slab, f1 is
friction between slab and ground, N1 is normal between slab and ground.

N N N1 N1

f f
(A) f (B) (C) 100N (D) 100N
f1 N f1 N
mg mg
block block Mg Mg
slab slab
PHYSICS/GR E-3/5
GUIDED REVISION
ENTHUSIAST & LEADER
Target : JEE(Main + Advanced) - 2025 COURSE
15. A block 'A' of mass 10 kg is placed on rough horizontal plank 'B' (x-y plane) where gravity act vertically
downward. µs = 1.2, µk = 1. (All units are in SI). Forces acting on block A is due to plank and earth only.
For the situation in column-I match the appropriate description in column-II.
z

A B

x mg
Column-I Column-II
 
ˆ V  3iˆ  4 ˆj, a  3iˆ  4 ˆj
(A) VA  3i, (P) Direction of friction on A is +ve y-direction.
B B
  
(B) VA  VB  3iˆ  4 ˆj, a B  8iˆ  6 ˆj (Q) Direction of friction on A is +ve x-direction.
  
(C) VA  VB  4ˆj, a B  8iˆ  6kˆ (R) Direction of acceleration of A is +ve y-direction.
    
(D) VA  VB  0, a B  8iˆ  6kˆ (S) aA  a B
(T) Friction force on A is neither in x-direction nor
y-direction but somewhere in between.

16. A particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential
component of its acceleration are equal. If its speed at t=0 is v0 then the time it takes to complete the first
R
revolution is 1  e  . Find the value of ()6.
v0
17. A small cubical block of mass 1 kg is placed inside a rough rectangular groove made in a circular rough
table as shown in the figure. Coefficient of friction for all the rough surfaces is  = 0.5. The table starts
rotating clockwise with angular acceleration 1 rad/sec2 in a horizontal plane about its axis. Find the time
(in sec) after which the block will start motion with respect to table. Assume the size of block slightly
smaller then the width of groove.

4m

3m

18. Figure shows a ball of mass m connected with two ideal springs of force constant k, kept in equilibrium
on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration
(in m/s2) of ball.
y

45°
45°

45°

E-4/5 PHYSICS/GR
GUIDED REVISION
ENTHUSIAST & LEADER
Target : JEE(Main + Advanced) - 2025 COURSE
19. Figure shows a block placed on a bracket. Bracket is placed unconstrained on a smooth floor, it is pulled

by a constant force F  6iˆ horizontally. Block is projected with velocity v0 relative to bracket as shown
in figure. If time in second after which it stops relative to bracket is t, then find the value of 0.6 t.
Horizontal surface of bracket is smooth while vertical surface is rough (Given : m = 1kg, M = 5kg,
v0 = 5m/s, µ = 0.5)

gh
z

Rou
v0 Smooth

m
M
F
x

20. There is a parabolic-shaped bridge across a river of width 100m. The highest point of bridge is 5m
above the level of the banks. A car of mass 1000 kg is crossing the bridge at a constant speed of
20 ms–1.Using the notation indicated in the figure, the force exerted on the bridge by the car when it is :
at the highest point of the bridge is (1200x) N. Find x (Ignore air resistance and take g as 10 ms–2)

 m
h

GR # (NLM, FRICTION AND CIRCULAR MOTION) ANSWER KEY


1. Ans. (A) 2. Ans. (C) 3. Ans. (D) 4. Ans. (A,B,C,D)
5. Ans. (B,C,D) 6. Ans. (A,B,C) 7. Ans. (B,C,D) 8. Ans. (A)
9. Ans. (A) 10. Ans. (A) 11. Ans. (A) 12. Ans. (C)
13. Ans. (C) 14. Ans. (C) 15. Ans. (A) (P, R) (B) (S, T) (C) (S,T) (D) (S, T)
16. Ans. 729 17. Ans. 2 18. Ans. 5 19. Ans. 5
20. Ans. 7

PHYSICS/GR E-5/5

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