1. PE, KE Examples Answers
1. A shotput has a mass of 7.0 kg. Find the potential energy
of a shotput raised to a height of 1.8 m.
m = 7.0 kg h = 1.8 m
PE = m g h
= ( 7.0 kg )( 9.8 m/s2 )( 1.8 m )
PE = 123 J

2. 2. If a car of mass 1000 kg is raised to a height of 2.5 m
by a hydraulic lift, what is its PE?
m = kg h = 2.5 m
PE = m g h
= ( 1000 kg )( 9.8 m/s2 )( 2.5 m )
PE = J

3. 3. A bowling ball of mass 7.0 kg moves at 6.5 m/s down a
lane. Find the kinetic energy of the ball.
m = 7.0 kg v = 6.5 m/s
KE = ½ m v2
= ½ ( 7.0 kg )( 6.5 m/s )2
KE = 148 J

4. 4. A car of mass 1200 kg moves at 24 m/s down a street.
Find the KE of the car.
m = kg v = 24 m/s
KE = ½ m v2
= ½ ( 1200 kg )( 24 m/s )2
KE = J
KE = J

5. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
2.0 kg
(a) How much work was done to lift
it to that height?
2.5 m
W = F d
F = force needed to lift ball
= weight of ball
= m g
d = distance raised
= height

6. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
2.0 kg
(a) How much work was done to lift
it to that height?
2.5 m
W = m g h
F = force needed to lift ball
= weight of ball
= m g
d = distance raised
= height

7. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
2.0 kg
(a) How much work was done to lift
it to that height?
2.5 m
W = m g h
= ( 2.0 kg )( 9.8 m/s2 )( 2.5 m )
W = 49 J

8. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
2.0 kg
(b) What is the PE of the ball at that height?
PE = m g h
2.5 m
= ( 2.0 kg )( 9.8 m/s2 )( 2.5 m )
PE = 49 J
This is the same as (a); the answer to (b) could also be
obtained by the Work-Energy Theorem (the energy
acquired by the ball = the work done on it)

9. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
PE = 49 J
2.0 kg
(c) What is the KE at that height?
KE = ½ m v2
2.5 m

10. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
PE = 49 J
2.0 kg
(c) What is the KE at that height?
KE = ½ m v2
2.5 m
v = 0
KE = 0

11. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
PE = 49 J
KE = 0
2.0 kg
(d) What is the Total Energy of the system?
By Conservation of Energy,
Total Energy = PE + KE
2.5 m

12. 5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there.
PE = 49 J
KE = 0
2.0 kg
(d) What is the Total Energy of the system?
By Conservation of Energy,
Total Energy = PE + KE
2.5 m
= 49 J + 0
Total Energy = 49 J

13. 5. (e) The ball is released and falls to the ground. At the
instant before the ball strikes the ground, what is
the PE of the ball?
PE = m g h
h = 0
PE = 0
2.0 kg

14. 5. (f) The ball is released and falls to the ground. At the
instant before the ball strikes the ground, what is
the KE of the ball?
Total Energy = PE + KE
49 J = KE
PE = 0
KE = 49 J
2.0 kg
Total Energy = 49 J
from (d)

15. 5. (g) What is the speed of the ball as it strikes the ground?
2 (
KE = ½ m v2
) 2
Solve for v
2 KE = m v2 m m
2 KE
v2 =
PE = 0 m
KE = 49 J
2.0 kg
2 KE
2 ( 49 J )
v = = m
2.0 kg
Challenge: Prove that J
v = 7.0 m/s
= m/s kg

16. 5. (h) As the ball is falling, PE is transformed into KE.
When the ball is 1.0 m above the ground, what is
the PE?
PE = m g h
= ( 2.0 kg )( 9.8 m/s2 )( 1.0 m )
2.0 kg
PE = J
1.0 m

17. 5. (i) What is the ball’s KE when it is 1.0 m above the ground?
Total Energy = PE + KE
PE = 19.6 J
49 J = J + KE
2.0 kg J J
1.0 m
KE = J
Total Energy = 49 J
from (d)

18. 5. (j) What is the ball’s speed when it is 1.0 m above the ground?
2 (
KE = ½ m v2
) 2
2 KE = m v2
PE = 19.6 J m m
KE = 29.4 J
2.0 kg
2 KE
1.0 m
v2 = m
2 KE
2 ( 29.4 J )
v = = m
2.0 kg
v = 5.4 m/s

19. 6. A spring that has a spring constant of 45.0 N/m is
compressed 23.0 cm.
(a) What force is required to compress the spring?
F = k x
k = N/m
x = cm
= m
F = ( 45.0 N/m )( 0.23 m )
F = N

20. 6. A spring that has a spring constant of 45.0 N/m is
compressed 23.0 cm.
(b) What is the elastic potential energy of the spring?
PEe = ½ k x2
k = N/m
x = cm
= m
PEe = ½ ( 45.0 N/m )( 0.23 m )2
PEe = J

21. 6. (c) A mass of 1.2 kg is placed on the spring and the
spring is released. What is the speed of the mass as it
leaves the spring?
PE = J
When spring is released, PE of
spring is transformed into KE
of the mass
1.2 kg
Mass leaves spring
with 1.19 J of KE
KE = ½ m v2
2 KE
2 ( 1.19 J )
v = =
= v = 1.4 m/s m
1.2 kg

22. 7. Consider : (a) Find the PE at A PE = m g h
200 kg
50 m
30 m
A B C
(a) Find the PE at A
PE = m g h
= ( 200 kg )( 9.8 m/s2 )( 50 m )
PE = J

23. (c) the Total Energy at A
7. Consider :
200 kg
50 m
30 m
A B C
(b) the KE at A
v = 0
KE = 0
(c) the Total Energy at A
Total Energy = PE + KE
= J + 0
Total Energy = J

24. 7. Consider : (d) the PE at B h = 0 PE = 0 (e) the KE at B
200 kg
50 m
30 m
A B C
(d) the PE at B
h = 0
PE = 0
(e) the KE at B
Total Energy = PE + KE
J = KE
KE = J

25. 7. Consider : (f) the speed at B 2 KE 2 ( 98 000 J ) v = = m 200 kg
A B C
(f) the speed at B
2 KE
2 ( J )
v = = m
200 kg
v = m/s

26. 7. Consider : (g) the PE at C PE = m g h
200 kg
50 m
30 m
A B C
(g) the PE at C
PE = m g h
= ( 200 kg )( 9.8 m/s2 )( 30 m )
PE = J

27. 7. Consider : (h) the KE at C Total Energy = PE + KE
200 kg
50 m
30 m
A B C
(h) the KE at C
Total Energy = PE + KE
J = J + KE J J
KE = J

28. 7. Consider : (i) the speed at C 2 KE 2 ( 39 200 J ) v = = m 200 kg
A B C
(i) the speed at C
2 KE
2 ( J )
v = = m
200 kg
v = m/s