9.3 Energy Solution A.

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Work & Energy Homework - ENERGY TRANSFORMATIONS WITH NO WORK OR FRICTIONAL LOSSES

Stragety for Working Energy Problems

1. Select the system and its starting (E1) and ending State (E2). Write the Law of Conservation of Energy (E1 = E2)
2. Write the Equation for each State, include all types of Energy (use the initials for the forms of energy)

E1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
3. Cross out values that are equal to zero to simplify the equation.
4. Substitute in the equations for the remaining forms of energy.

1 mv 2  mgh  1 kx2  1 mv 2  mgh  1 kx2

2 1 1 2 1 2 2 2 2 2
5. Solve the remaining equation for the desired variable.
6. Plug and Chug

Law of Conservation of Energy – Use the Law of Conservation of Energy to solve the following problems.
1. An elevator has an unknown mass. The elevator is at the top floor of the Sears Tower that is 440 meters high. If the
cable holding the elevator were to snap, what velocity would it have when it hits the ground floor? (We must assume
no friction is present)

The mass is not given so it must cancel.

h1 = 440 meters
acceleration due to gravity = 9.8 m/s2 ≈ 10 m/s2
h2 = 0 meters

All the energy in State 1 is gravitational energy; so all others can be cancelled. All the energy in State 2 is
in the form of kinetic energy so all others can be cancelled.

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 No spring is involved so EPE is zero in all States

 The object is not moving in State 1 so KE1 = zero
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0
 The object is at its highest velocity as it hits the ground so it does have Kinetic Energy in State 2 – It is
important to understand that we are looking for the velocity with which it hits the ground.

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

 { GPE1 } = { KE2 }

 
mgh1  1 mv22
2

solving this for velocity gives

 
v2  2 gh

  m
v2  210 2 440m  94m / s
 s 
Plugging in the appropriate values gives:

2. A general form of the above question – An object with a mass of m starts at rest a height h above the ground. At
what velocity will it hit the ground?

All the energy in State 1 is gravitational energy, so all others can be cancelled. All the energy in
State 2 is in the form of kinetic energy so all others can be cancelled.

E1 = E2

{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

 
mgh1  1 mv22
2

solving this for velocity gives

 
v2  2 gh
3. Question 2 is an example of Gravitational Potential Energy being converted to Kinetic Energy.

4. A baseball with a mass .145 kg is thrown with a velocity 10 m/s straight down. If the ball starts from a height of 20
meters, how fast is the ball going when it hits the ground?

LIST VARIABLES

m = .145 kg
h1 = 20 meters
v1 = 10 m/s
acceleration due to gravity = 9.8 m/s2 ≈ 10 m/s2
h2 = 0 meters
v2 = ??
 All the energy in State 1 is Gravitational energy & Kinetic. There is not spring so Elastic Potential Energy
is zero for all cases. All the energy in State 2 is in the form of Kinetic Energy so all others can be
cancelled.

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 No spring is involved so EPE is zero in all States

 The object is moving in State 1 so KE1 ≠ zero so it does not cancel out this time.
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0
 The object is at its highest velocity as it hits the ground so it does have Kinetic Energy in State 2 – It is
important to understand that we are looking for the velocity with which it hits the ground.

0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ KE1 +GPE1 } = { KE2 }

1 mv 2  mgh  1 mv 2

2 1 1 2 2

solving this for velocity gives

  
v2  v12  2 gh
Plugging in the appropriate values gives:

 m m m
v2  (10m / s)2  2(10m / s 2 )(20m)  500  10 5  22.3
s s s

5. A baseball with a mass .145 kg is thrown with a velocity 10 m/s straight up. If the ball starts from a height of 20
meters, how fast is the ball going when it hits the ground? (See the solution to number 7)

6. A baseball with a mass .145 kg is thrown with a velocity 10 m/s 45˚ above the horizontal. If the ball starts from a
height of 20 meters, how fast is the ball going when it hits the ground? (See the solution to number 7)

7. A baseball with a mass .145 kg is thrown with a velocity 10 m/s 45˚ below the horizontal. If the ball starts from a
height of 20 meters, how fast is the ball going when it hits the ground?
 Upon reviewing the solution of the previous questions you will see that angle is not part of any
of the equations. Notice especially the equation for Kinetic Energy, which does not take any
angle. This means that the Kinetic Energy is Independent of the angle at which the object is
traveling. Or in common terms “Angle don’t matter”. Your so SILLY if you thought angle
mattered and you did lots of work to get through the last few questions…. I am laughing just a
bit.

So you will see that analyzing the equations can save you a lot of work!!

8. Questions 3-7 are examples of Kinetic Energy & Gravitational Potential Energy- being converted to
Kinetic Energy.

9. A ball is thrown from the ground (assume it leaves the person’s hand at zero elevation) straight up with a velocity of
15 m/s. What is the greatest height the ball reaches?

LIST VARIABLES

m = not given so it likely cancels out

h1 = 0 meters
v1 = 15 m/s
acceleration due to gravity = 9.8 m/s2 ≈ 10 m/s2
h2 = ?? meters
v2 = 0

All the energy in State 1 is Kinetic Energy. There is not spring so Elastic Potential Energy is zero for all
cases. All the energy in State 2 is in the form of Gravitational Potential Energy so all others can be
cancelled.

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 No spring is involved so EPE is zero in all States

 The object is moving in State 1 so KE1 ≠ zero .
 The object has no elevation in State 1 so GPE1 is zero (it does cancel)
 The object has some elevation in State 2 so GPE2 ≠ 0 (this does not cancel)
 The object changes all of its Kinetic Energy to Gravitational Potential Energy

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ KE1 } = { GPE2 }
 1 mv 2   mgh
2 1 1

solving this for height gives


1 v12
h  
2 g
Plugging in the appropriate values gives:


1 v12 1 15m / s 
2
h     2
 11.25m
2 g 2 10m / s

10. A ball is thrown from the ground (assume it leaves the person’s hand at zero elevation) straight up with a velocity of
v. What is the greatest height the ball reaches?

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ KE1 } = { GPE2 }

1 mv 2   mgh
2 1 1

solving this for height gives


1 v12
h  
2 g
11. Questions 9&10 are examples of Kinetic Energy - being converted to Gravitational Potential Energy.
 12. A baseball at rest is dropped from a height of 10 meters onto a spring with a spring constant of 5N/m. How much
will the spring be compressed?

m = .145 kg from the other questions

h1 = 10 meters
Acceleration due to gravity = 9.8 m/s2 ≈ 10 m/s2
h2 = 0 meters
k = spring constant = 5 N/m
x = compression of spring = ?? m

All the energy in State 1 is gravitational energy, so all others can be cancelled. All the energy in State 2 is
in the form of Elastic Potential Energy in the Spring so all others can be cancelled.

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 The spring is assumed not to be compressed in State 1 so EPE1 = 0 and can be cancelled
 The object is not moving in State 1 so KE1 = zero and can be cancelled
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0, this can be cancelled
 At the bottom of the journey of the ball it has compressed the spring as far as it will be able, so the
energy has all be converted into Elastic Potential Energy in the spring.

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ GPE1 } = { EPE2 }


mgh1  1 kx2
2
Notice that this time the mass does not cancel, as it did in prior problems.
Solving this equation of x, the compression of the spring gives:

2mgh1
x
k
Substituting in the appropriate values gives:
 
2mgh1 2  (.145kg)  (10m / s 2 )  (10m)
  2.4m
k 5N / m
13. An object with a mass of m at rest is dropped from a height of h onto a spring with a spring with a spring constant of
k. How much will the spring be compressed?

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 The spring is assumed not to be compressed in State 1 so EPE1 = 0 and can be cancelled
 The object is not moving in State 1 so KE1 = zero and can be cancelled
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0, this can be cancelled
 At the bottom of the journey of the ball it has compressed the spring as far as it will be able, so the
energy has all be converted into Elastic Potential Energy in the spring.

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ GPE1 } = { EPE2 }


mgh1  1 kx2
2
Notice that this time the mass does not cancel, as it did in prior problems.
Solving this equation of x, the compression of the spring gives:

2mgh1
x
k

14. Questions 12&13 are examples of Gravitational Potential Energy - being converted to Elastic Potential
Energy.
 15. A baseball traveling at 2 m/s straight down is at a height of 10 meters and falls onto a spring with a spring with a
spring constant of 5N/m. How much will the spring be compressed?

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 The spring is assumed not to be compressed in State 1 so EPE1 = 0 and can be cancelled
 The object is not moving in State 1 so KE1 = zero and can be cancelled
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0, this can be cancelled
 At the bottom of the journey of the ball it has compressed the spring as far as it will be able, so the
energy has all be converted into Elastic Potential Energy in the spring.

0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ KE1 + GPE1 } = { EPE2 }

1 mv 2  mgh  1 kx2

2 1 1 2
Notice that this time the mass does not cancel, as it did in prior problems.
Solving this equation of x, the compression of the spring gives:
 
m(v12  2 gh1 )
x
k

Plugging in the appropriate values gives:

 
m(v12  2 gh1 ) (.145kg){(2m / s) 2  2(10m / s 2 )(10m)}
x   2.43m
k 5N / m

16. A baseball traveling at 2 m/s straight up is at a height of 10 meters and falls onto a spring with a spring with a
spring constant of 5N/m. How much will the spring be compressed?
 It does not matter that the ball is going a different direction in this case. As with the previous problems (4-
7) the direction does not effect the total energy of the object. The Energy in State one is only related to the
velocity, mass and height of the object.

17. An object traveling v straight down is at a height of m and falls onto a spring with a spring with a spring constant
of k. How much will the spring be compressed?

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 The spring is assumed not to be compressed in State 1 so EPE1 = 0 and can be cancelled
 The object is not moving in State 1 so KE1 = zero and can be cancelled
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0, this can be cancelled
 At the bottom of the journey of the ball it has compressed the spring as far as it will be able, so the
energy has all be converted into Elastic Potential Energy in the spring.

0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ KE1 + GPE1 } = { EPE2 }

1 mv 2  mgh  1 kx2

2 1 1 2
Notice that this time the mass does not cancel, as it did in prior problems.
Solving this equation of x, the compression of the spring gives:
 
m(v12  2 gh1 )
x
k

18. Questions 15&17 are examples of Gravitational Potential Energy & Kinetic Energy - being converted to
Elastic Potential Energy..
Practical Problems Employing the Law of Conservation of Energy – Use the Law of Conservation of Energy
to solve the following problems.

19. A roller coaster has a max height of 40 meters above its lowest point (about 120 feet)
assuming there is no friction, what is the velocity of the roller coaster at the bottom of
the hill?

The mass is not given so it must cancel.

h1 = 440 meters
acceleration due to gravity = 9.8 m/s2 ≈ 10 m/s2
h2 = 0 meters

All the energy in State 1 is gravitational energy, so all others can be cancelled. All the energy in State 2 is
in the form of kinetic energy so all others can be cancelled.

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }
Analyze the States to cancel things out:

 No spring is involved so EPE is zero in all States

 The object is not moving in State 1 so KE1 = zero
 The object has elevation in State 1 so GPE1 is not zero (it does not cancel)
 The object has no elevation in State 2 so GPE2 = 0
 The object is at its highest velocity as it hits the ground so it does have Kinetic Energy in State 2 – It is
important to understand that we are looking for the velocity with which it hits the ground.

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ GPE1 } = { KE2 }

 
mgh1  1 mv22
2

solving this for velocity gives

 
v2  2 gh

  m
v2  210 2 40m  28.3m / s
 s 
Plugging in the appropriate values gives:
20. If the roller coaster starts at a height of 40 meters and a velocity of zero, how high a hill will the roller coaster be
able to climb?

The GPE that the roller coaster starts with turns into Kinetic Energy at the bottom of the hill and then back into GPE
while going up the next hill. So GPE1 = GPE2 and since mass is constant the height is starts with is the height it will
end up with.

E 1 = E2

{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ GPE1 } = { GPE2 }
 
mgh1  mgh2
 
mgh1  mgh2
h1 = h 2
If there is friction of course some energy will be lost and the roller coaster will not be able to reach the original
height. In the lack of friction the roller coaster should be able to reach its original height.

21. If the roller coaster started from twice as high, would it be going twice as fast at the bottom of the hill? Calculate the
factor by which velocity at the bottom of the hill is increased if the starting height is increased by a factor of 2.

Wouldn’t it be nice if it were this simple? Your intuition will tell you it is this simple but the
reality is that it is not.
 
From above v2  2 gh

Create a ratio

 
v2 2 g 2h1
  
v1 2 gh1
solve for the relation between the velocities

v2
  2
v1
22. Tarzan finds himself stuck on the opposite side of a river filled with
hungry crocodiles. To make it up to Jane there is a vine upon which
he can swing up to her. The vine is 10 meters long and starts out
hanging straight down as shown by the dashed line. To swing up to
Jane, Tarzan runs and jumps onto the vine. He must swing the vine
through an angle ( θ ) of 40˚ to make it up to the other side. How HYP
fast must he be going to make it up to Jane. (Assume he hits the rope ADJ
with a horizontal velocity)

From the length of the vine and the angle you can find how much
Tarzan increases his height. Create and label adiagram of the
situation. From the angle and length of the vine you may calculate
the ADJ side of the triangle. The amount Tarzan is lifted is equal to
the length of the vine minus the ADJ side.

Amount Tarzan is Lifted, h = (Length of Vine) – (ADJ)

h = (10 m) – (10 m cos 40) = 2.4 meters

Now that the change in height is known, the problem is reduced to Kinetic Energy is being turned into Gravitational
Potential Energy.

E 1 = E2
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ KE1 } = { GPE2 }

1 mv 2   mgh
2 1 1

solving this for velocity gives:

 
v1  2 gh

Plugging in the appropriate values gives:

6.9 m/s
 23. The launch mechanism of a spring gun has a spring that can be compressed .12 meters.
The bullet (20 grams) reaches a maximum height of 20 meters. (A) Calculate the spring
constant of the spring. (B) Calculate the velocity of the bullet as it leaves the spring.

(A) Here the Spring Energy (EPE) is being changed into (GPE).

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ EPE1 } = { GPE2 }

1 kx2  mgh
2 2

solving for the spring constant gives:


2mgh
k
x2
Plugging in gives: 544.4 N/m

(B) Calculate the velocity of the bullet as it leaves the spring.

There are two ways to find this solution. 1) use the height that the bullet reaches (GPE) to find the
velocity (KE) it must have been going when it left the gun and 2) use the spring energy (EPE) to find the
velocity it must have had. Neither is better.

Method 1 Method 2

{ KE1 } = { GPE2 } { EPE1 } = { KE2 }

1 2 
mv1  mgh2 1 kx2  1 mv22
2 2 2

solving for the velocity gives: solving for the velocity gives:
 
v1  2 gh2  kx2
v
m
Plugging in gives: Plugging in gives:

 
v1  2 9.8m / s 2 20m  19.8m / s  544.4 N / m.12m2
v1   19.8m / s
.02kg
 24. The length of a pinball machine tilted surface is 1.2 meters and is angled at 10˚ . If the spring can be compressed by
a distance of 5 cm, what must the spring constant be so that the ball will make it to the top of the sloped surface?

The length of the pinball machine in the HYPotenuese of the triangle (1.2 meters). The height the ball can reach is the
OPPosite side. First find the OPPosite side.

OPP  HYP  sin   1.2msin10  .21 meters

OPP
sin   so
HYP
The GPE the ball obtains comes from the spring (EPE) so Elastic Potential Energy is turned into Gravitational Potential
Energy.

0 0 0 0
{ KE1 + GPE1 + EPE1 } = { KE2 + GPE2 + EPE2 }

this leaves the following:

{ EPE1 } = { GPE2 }

1 kx2  mgh
2 2

solving for the spring constant gives:


2mgh
k
x2
where

m = this is not given so the final solution will be an equation including the variable m for mass
g = the gravitational acceleration
h = the height the ball needs to reach = .21 m
x = the amount the spring can be compressed = .05 m

Plugging in gives:

2mgh 2  m  9.8  .21
k   1646m
x2 .052
A ballistic pendulum is a simple device used to indirectly measure the velocity of a bullet. The pendulum is made of a
substance that allows the bullet to become imbedded in it. The pendulum catches the bullet and the pendulum is moved and
swung upwards turning the kinetic energy into gravitational potential energy. Given that the mass of the bullet is m, the mass
of the pendulum is M and the pendulum is lifted a distance y; derive and equation that will give the velocity v of the bullet as
a function of the given values.

THE BALLISTIC PENDULUM

Ballistic Pendulum Just Prior to Ballistic Pendulum at Maximum

Bullet Strike - a mass M hangs from Height – the combined mass (M +m)
Ballistic Pendulum at two lines is about to be struck by a swings to its maximum height above
Rest - a mass M hangs bullet of mass m traveling horizontally the rest position y.
from two lines. with a velocity v. The combined mass
will swing as shown

25. You are asked to design a ballistic pendulum for a gun that fires bullets from 600 ft/s to 800 ft/s. The masses of the
bullets range from 45 grams to 47 grams. The ballistic pendulum must not be lifted more than half a meter. What
should its mass be?

26. You are asked to design a ballistic pendulum for a gun that fires bullets from 600 ft/s to 800 ft/s. The masses of the
bullets range from 45 grams to 47 grams. The ballistic pendulum pivot lines must not be rotate more than 45˚. What
should its mass be? How long should the support lines be?

27. What is the Lowest Energy Theorem?