Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.

08 Work, Energy and Power gravitational PE when they reach the ends of
the tracks, and hence by conservation of
Checkpoint
energy, both gain the same amount of KE .
Checkpoint 1 (p.194)
(b) B
1. (a) Yes As Y slides along a steeper slope, its
(b) No. The displacement of the bag is gravitational PE is converted into KE faster
perpendicular to the force exerted on it. than X does. Hence, Y gains its speed faster
(c) Yes and reaches the end first.

2. Work done = F s cos θ = (50)(30)cos 30° ≈ 1300 J. 3. (a) Work done = mgh = 500 × (1000 − 500)
= 2.5×105 J.
3. (a) The component of the weight along the slope
F∥ = 10 sin 30° = 5 N. (b) (gravitational) PE of the skydiver →
internal energy of the air and the skydiver
(b) Work done by the weight
W = F∥s = 5 × 1 = 5 J.
Checkpoint 4 (p.218)
(c) Since the block slides down at constant
speed, the work done by the weight is used 1. By conservation of energy,
as the work done against friction. So, the gain in PE = loss in KE
work done against friction is 5 J. mgh = mv2

Checkpoint 2 (p.204) ∴ h= = = 3.26 m

1. (a) The ball rises 3.26 m.

speed v / m s−1 kinetic energy / J 2. By conservation of energy,

1 0.005
the work done by the resistive force = loss in
(gravitational) PE
2 0.02
= mgh =5000 × (20 + 2) = 1.1 × 105 J
3 0.045
(b) Average resistive force

height h / m gravitational PE / J F= = = 55000 N

0 0
0.5 0.0491
Checkpoint 5 (p.224)
1.0 0.0981
1.5 0.147 1. Work done by the crane
2. W = F∥s = 20000 × 20 = 4 × 105 J

 The net force on the block = 50 − 30 = 20 N. Power of the crane

Hence, the block accelerates upwards. P= = = 8000 W
(a) The gravitational PE of the block increases.
Alternative Solution:
(b) The KE of the block increases.
Speed of the load
(c) The mechanical energy of the block
v= = = 0.4 m s−1
increases.
Power of the crane
(d) The total work done on the block increases.
P = Fv = 20000 × 0.4 = 8000 W
2. B
Checkpoint 3 (p.208)
Let a be the acceleration of the car.
1. Power of the car
position gravitational PE / J KE / J
P= = = =
A 300 20
B 120 200
C 120 200
D -30 350
2. (a) C
The marbles lose the same amount of
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.2

Exercise
2. C
Exercise 8.1 (p.194)
By KE = mv2, we have KE ∝ v2. If v is doubled,
1. B
then the KE is 4 times of the original value.
W = F s = mas
⇒ [ J ] = [kg][m s−2][m] = [kg][m2 s−2] 3. A

2. B Since KE = mv2, the slope of the graph is . If

If the length of the barrel is longer, the cannonball the mass m is doubled, then the slope is doubled.
is acted on by the force for a longer distance (i.e.
4. C
larger s). By W = F s, a larger s leads to a larger
The change in the height of the marble is the same
work W to be done on the cannonball, which hence
no matter it slides down rails A or B.
gains a larger KE.
5. D
3. B
Note that work has to be done to stretch the spring.
4. A The spring does not gain any gravitational PE
The normal reaction exerted by the ground is since it (more precisely, its c.g.) does not have any
perpendicular to the direction of motion of the vertical motion.
crate.
6. (a) The KE of the car is
5. The net work done = Fnet s = (5 − 2) × 3 = 9 J.
KE = mv2
6. (a) The work done by the pulling force
= F∥s = (20 cos 45°)(50) = 707.1 ≈ 707 J. = (2000)(122) = 1.44 × 105 J

(b) The work done against friction (b) When the car comes to a stop, the KE of the
= f s = (8)(50) = 400 J. car must have been transformed into the
(c) The energy gained by the suitcase work done against friction. Given that the
= 707.1 – 400 ≈ 307 J. braking force is 3500 N, we have
KE = F s
7. By W = F s, we have
1.44 × 105 = 3500s
energy gained = area under the F–s graph
∴ s ≈ 41.1 m < 50 m
= =5J
Yes. The car can stop before the flock of
So the energy gained by the block is 5 J. sheep.
8. (a) Since the dumbbell is lifted at a constant 7. The gravitational PE of John is given by
speed, the force exerted by Carol is equal to gravitational PE = mgh
the weight of the dumbbell. = (70)(9.81)(6 sin 38°)
The work done by Carol is ≈ 2540 J
W = F s = mgs = (10)(9.81)(0.3) ≈ 29.4 J 8. (a) If the cannonball is fired horizontally, the
(b) When the dumbbell falls, the only force work done on the cannonball is equal to the
acting on the dumbbell is the gravitational gain in its KE, and hence
force exerted by the Earth. Fs= mv2
The work done by the Earth is
W = F s = mgs = (10)(9.81)(0.8) ≈ 78.5 J (20000)(1) = (15)v2
9. (a) The work done against friction is ∴ v ≈ 51.6 m s−1
W = f s = (0.3)(2π × 0.2 × ) ≈ 0.0942 J The speed of the cannonball is 51.6 m s−1.
(b) Free body diagram of the cannonball when it
(b) The work done by the gravitational force is
reaches the end of the barrel:
W = mgs = (0.1 × 9.81)(0.2) ≈ 0.196 J

Exercise 8.2 (p.204)

1. (a) The gain in PE
= mgh = (120)(9.81)(2) ≈ 2350 J.
(b) The KE
= mv2 = (0.43)(82) ≈ 13.8 J. Since the net work done on the cannonball is
equal to the gain in its KE , we have
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.3

(F − mg)s = mv2 3. D
4. B
(20000 – 15 × 9.81)(1) = (15)v2
PE = mgh ⇒ PE ∝ h.
∴ v ≈ 51.4 m s−1 The PE of the object increases linearly with its
The speed of the cannonball is 51.4 m s−1. height h. Since the mechanical energy of the object
is conserved, its KE decreases linearly with the
OR: The work done on the cannonball is
height h.
equal to the sum of its KE and gravitational
PE, and so 5. His idea is impractical. According to the law of
conservation of energy, no device, including the
Fs= mv2 + mgh
torch, can create energy. That is why the torch can
(20000)(1) = (15)v2 + (15)(9.81)(1) never output more energy than the energy input.

∴ v ≈ 51.4 m s−1 6. (a) When the block falls in the air, its
gravitational PE is converted to its KE. When
9. The y–intercept of the graph represents the initial it presses against the spring, both its KE and
gravitational PE of the block. Thus gravitational PE are converted to the elastic
grav. PE = mgh PE of the spring.
1000 = (10)(9.81)h
(b) The elastic PE gain = the gravitational PE
∴ h = 10.19 ≈ 10.2 m
loss
The vertical distance travelled by the block is
= mgh = (1)(9.81)(1 + 0.08) ≈ 10.6 J.
10.2 m.
The elastic PE stored in the spring when the
Consider the figure below.
block is at the lowest position is 10.6 J.
7. (a) By conservation of energy,
KE gain = gravitational PE loss
=

Therefore, we have The speed of the skateboarder when he is at

the end of the ramp is
sin θ = ⇒ θ ≈ 11.8°
v= = ≈ 6.26 m s−1
10. (a) Take the initial gravitational PE of the box as
(b) The speed of the heavier skateboarder will
zero. The work done on the box is
be the same as the lighter skateboarder,
W = F s = (8)(20) = 160 J
because the speed at the end of the ramp is
(b) The gain in KE is independent of its mass.
KE = mv2 = m(22) = 2m
8. (a) Loss in KE = (m)(u2 − v2)
The gain in gravitational PE is
= (1500)(252 − 52)
PE = mgh = m(9.81)(20 sin 30°) = 98.1m
= 4.5 × 105 J
Combining the equations, we have
work done = KE gain + PE gain (b) Gain in PE = mgh
160 = 2m + 98.1m = (1500)(9.81)(25)
= 3.679 × 105 = 3.68 × 105 J
∴ m= ≈ 1.60 kg
(c) By conservation of energy, we have
KE loss = PE gain + work done
Exercise 8.3 (p.218) 4.5 × 105 = (3.679 × 105) + W
1. C ∴ W = 8.213 × 104 J
Without friction, the mechanical energy of the By W = F s, we have
child is conserved. 8.213 × 104 = f (50 + 25 / sin 50°)
2. C ∴ f ≈ 994 N
Even though the rod changes the motion of the Therefore, the friction acting on the roller
string and the bob, the system still obeys the law coaster is 994 N.
of conservation of energy. Unless there is energy
lost to the surroundings, the bob will finally
returns to the same height at R.
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.4

Alternative Solution: ∴ h' = 0.7h = 0.7 × 100 = 70 m

We can also find the friction f using the He is 70 + 20 = 90 m above the river.
equations of motion. Let C be the lowest (c) The energy is lost to the surroundings in the
point of the slope. form of heat and sound.
From A to C: Net force = friction f. 11. (a) By conservation of energy, the mechanical
So, acceleration a = −f/m. energy loss of the weight (of mass M) is
equal to the energy gain of the water (of
By , mass m). So, we have
Mgh = mc∆T
(6)(9.81)(0.8) = (0.5)(4200)∆T
From C to B: Net force = mg sin 50° + f. ∴ ∆T ≈ 0.0224 °C
So, acceleration The water temperature increases by
a = −(net force)/m = −g sin 50° − f/m 0.0224 °C.
Again, by , (b) The energy is lost to overcome the friction
between the rope and the pulley.
12. (a) By conservation of energy, the loss in PE
Subsituting by the expression above and
equals the gain in KE. Hence, we have
solving gives f = 994 N.
KE = mgh = (70)(9.81)(10 − 3)
9. (a) The initial height of the bob above the lowest = 4807 ≈ 4810 J
point is The KE of David is 4810 J when he just
h = 1 – cos 30° = 0.1340 m reaches the cushion.
(b) Take the gravitational PE of the bob at the (b) By conservation of energy, the sum of the
lowest position as zero. By conservation of losses in KE and PE is equal to the work done
energy, we get by the cushion. Hence, we have
initial PE + initial KE = final KE mgh = f s
+ = (70)(9.81)(10 − 1) = f (2)
∴ f ≈ 3090 N
Rearranging the equation, we have Therefore, the average force acting on David
v= = by the cushion is 3090 N.
= 1.9049 ≈ 1.90 m s−1 Alternative Solution:
When the bob is given a slight push, its speed Take the downward direction as positive. By
at the lowest position is 1.90 m s−1.
KE = mv2, the speed of David when he just
(c) By conservation of energy, we have
reaches the cushion is
KE loss = PE gain
u= = = 11.72 ms-1
=
By v2 − u2 = 2as, we have
Rearranging the equation, we have
0 − 11.722 = 2a(2) ⇒ a = −34.34 m s−2
h= = ≈ 0.185 m The average deceleration of David during the
impact with the cushion is 34.34 m s−2. By
The bob reaches a height of 0.185 m on the
Newton’s second law, we have
other side of the pendulum.
mg − f = ma
10. (a) The gravitational PE loss (70)(9.81) − f = (70)(−34.34)
= mg∆h = (70)(9.81)(120 − 20) = 68670 J. ∴ f ≈ 3090 N
(b) Suppose the man bounces up to the highest So the average force acting on David by the
position of height h' above his lowest cushion is 3090 N.
position. 13. By conservation of energy, we have
By conservation of energy, the gravitational PE loss = KE gain + work done
PE loss in the jump is equal to the
mgh = mv2 + W
gravitational PE gain after the bounce plus
the energy lost in the cord. Mathematically, (60)(9.81)(800) = (60)v2 + (432 × 103)
we have
= +
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.5

∴ v = 36 m s−1 7. (a) The gain in gravitational PE of each package

The terminal speed of the skydiver is 36 m s−1. is
E = mgh = (2)(9.81)(10 sin 20°)
14. (a) (i) KE of the athlete → elastic PE of the
= 67.10 ≈ 67.1 J
pole + gravitational PE of the athlete
(b) Since the speed of the package is constant
(ii) Elastic PE of the pole → gravitational
throughout the motion, its gain in KE is 0.
PE of the athlete
(c) The power of the conveyor belt is
(b) Assume that no energy is lost to the
surroundings. By conservation of energy, we P= = ≈ 11.2 W
have 8. Let n be the maximum number of passengers that
= the lift can carry each time.

Rearranging the equation, we have The gravitational PE gain of the passengers is

E = mgh = (70n)(9.81)(30) = 20601n
h= = ≈ 4.60 m
The gravitational PE gain of the lift is
The maximum the athlete can reach is E = Mgh = (1000)(9.81)(30) = 294300 J
(4.60 + 1) = 5.60 m. The lift provides the energy to lift itself and the
passengers to the 10th floor. By P = , we have
Exercise 8.4 (p.224)
60 × 103 = ⇒ n = 9.014
1. D
kW h is a unit of energy. Therefore, the lift can carry a maximum number of
2. C 9 passengers each time.
9. (a) For process 1, the energy transferred is
By W = F s cos θ and P = , we have
E1 = mv2 = (0.008)(3502) = 490 J
P= = (F cos θ) = Fv cos θ
For process 2, the energy transferred is
3. C
E2 = mv2
The work done on the box is

W = Fs = = Fat2 = (1500) = 5.787 × 105 J > E1

The power developed by the force is So the car transfers more energy.
P= =( )t = ( )t (b) For process 1, the average power is
Since both F and m are constants, a graph of P P1 = = = 8.167 × 105 W
against t is a straight line passing through the
origin. For process 2, the average power is

4. The power developed by Maggie is P2 = =

P= = ≈ 196 W = 1.157 × 105 W < P1

So the pistol has a larger average power.
5. The gain in gravitational PE by the volcanic ash is
10. The displacement of the car is given by the area
E = mgh = (3 × 109)(9.81)(20000)
under the graph, and hence
= 5.886 × 1014 J
The power of the eruption is s= = 60 m

P= = = 9.81 × 1012 W The power developed by the car is

6. The carrier travels at a constant velocity of P= = = 15000 W

45 km h−1 = = 12.5 m s−1

Chapter Exercise
By P = Fv, the driving force of the carrier is
Multiple-choice Questions (p.227)
F= = = 1.2 × 107 N
1. A
As the net force acting on the carrier is zero, the Statement (3) is incorrect. The track has friction,
water resistance acting on it is also 1.2×107 N. and so extra work has to be done to overcome it.
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.6

2. B (0.02)(4802) = (10000)(0.05n)
Note that the tram moves at a steady speed.
∴ n = 4.608
3. D
Thus, the bullet will be embedded in the 5th block.
The steel ball moves at its terminal speed, and so
its speed remains unchanged. For an object 10. D
moving at a constant speed, s ∝ t. Hence, its Option A is incorrect. If the block slides along an
gravitational PE decreases linearly with time. incline with constant friction, the mechanical
energy is not conserved.
4. C
Falling in midair: gravitational PE → KE Option B is incorrect. If the block falls at a
Bouncing at the racket: KE → elastic PE (by constant velocity, its KE remains unchanged.

changing its shape) Option C is incorrect. If the block accelerates on a

Bouncing off the racket: elastic PE → KE flat surface, its PE remains unchanged.

5. D 11. A
Statement (1) is correct. By Fnet = ma, if the mass m Since friction is negligible, all of the work W done
is doubled, the net force Fnet needed to reach the on the mass becomes the KE EK of the mass.
same acceleration a is doubled. Hence, the driving Hence, EK increases linearly with W.
force needed by the car is doubled. 12. A
Statement (3) is incorrect. Note that the ball has
Statement (2) is correct. By KE = mv2, if the
KE at P, and so it is possible for the ball to reach T
mass m is doubled, the KE required is doubled to if the KE is greater than the change in PE from P to
reach the same speed v. T.

Statement (3) is correct. By P = , since the 13. B

The output power of the motor is
acceleration lasts for the same time interval t for
both cars and the energy needed by the massive P= = = ≈ 8.2 W
car is doubled, the average input power required
14. D
by the massive car is doubled.
As both blocks slide down along frictionless
6. C surfaces for the same height H, all the gravitational
Let EK be the initial KE. When the speed of the PE they have lost are converted into KE, and so
marble is decreased by half, we have they have the same speed at the bottom of the
EK ' = m( )2 = ( mv2) = EK inclined plane, i.e. v1 = v2.
The distance travelled by the block on the incline
By conservation of energy, the gravitational PE =
planes is
. Hence, the ratio is 1 : 3.
s= ⇒ t=
7. B
In Fig. Q14(1), the sliding distance is shorter and
The bob has vertical displacement due to the
the acceleration is greater, and hence we can
weight W, and so it does a positive work.
conclude that t1 < t2.
The bob always moves in a direction
perpendicular to the tension T, and so it does not
Structured Questions (p.229)
work.
15. (a) The speed of the bead is maximum when it is
8. D
at C. By conservation of energy, we have
The feather reaches the terminal speed because
air resistance does a negative work on it and =
induces a loss in energy. By conservation of energy,
v2 = (9.81)(0.14) (1M)
we have
PE loss = KE gain + energy loss ∴ v ≈ 1.66 m s-1
⇒ PE loss > KE gain The maximum speed of the bead throughout
9. C the motion is 1.66 m s−1. (1A)
Assume that the bullet passes through n blocks. By (b) Let h be the vertical distance between A and
conservation of energy, D.
mv2 = F s =
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.7

(12) = 9.81h (1M)

(ii) By Newton’s second law, we have
Fnet = ma
∴ h = 0.05097 ≈ 0.0510 m 90000 = 20000a (1M)
D is 0.0510 m below A. (1A)
a = 4.5 m s−2 < 32 m s−2 (1M)
(c) Since E is (0.08 − 0.05097) = 0.02903 cm Hence, the jet fighter cannot take off on
above A, where the KE of the bead is zero, the aircraft carrier successfully.
the bead cannot pass E. (1A) (iii) Given that the jet fighter needs to reach
By conservation of energy, we have 80 m s−1 to take off.
= E= mv2 = (20000)(802)

(v2) = (9.81)(0.02903) (1M) = 6.4 × 107 J

The energy needed for the jet fighter to
∴ v ≈ 0.755 m s-1
take off is 6.4 × 107 J. (1M)
The bead must be projected at A with a
The total power developed during
speed of at least 0.755 m s−1 in order to pass
take-off is
E. (1A)
P= = = 3.2 × 107 W
(d) The bead must be projected at a higher
(1M+1A)
speed. (1A)
The friction exerted by the rail will take (b) When the jet fighter is caught by the cable,
energy away from the bead, i.e. work has to its remaining KE is converted to the elastic
be done to overcome the friction. Therefore, PE of the springs attached to the cable. (1A)
the bead has to possess more KE at A in 18. (a) (i) Energy can change from one form to
order to pass E. (1A) another, but it cannot be created or
16. (a) (i) When Sandy freewheels down the destroyed. (2A)
slope, her gravitational PE is converted (ii) When the hammer falls, its
into KE and thus she gains speed. (1A) gravitational PE is converted into its KE.
(1A)
(ii) By conservation of energy, we have
When the hammer hits the baseplate
= and drives it deeper into the ground,
the KE of the hammer is converted into
(9.81)(10 sin 30°) = v2 (1M)
the KE of the baseplate, but some KE
∴ v ≈ 9.90 m s-1 may be lost to the surroundings after
After travelling for 10 m, her speed is converting into internal and sound
9.90 m s−1. (1A) energy. (1A)
(b) (i) In the whole journey, the gravitational (b) (i) The KE of the hammer is given by
PE lost by Sandy is equal to the KE KE = mv2 (1M)
gained by her plus the work done by
the braking force f on the level road. So = (1500)(82) = 48000 J (1A)

we have
(ii) By conservation of energy, the initial
mgh = mv2 + f s
PE of the hammer equals its KE just
(75)(9.81)(20 sin 30°) before it hits the baseplate. (1A)
= (75)(52) + 10f (1M)
Hence, the initial PE of the hammer is
48000 J. (1A)
∴ f = 642 N
(iii) Considering the gravitational PE of the
So the average braking force is 642 N.
(1A)
hammer, we have
PE = mgh (1M)
(ii) Part of the PE is converted into the
48000 = (1500)(10)h
internal energy of the brake pads. (1A)
∴ h = 3.2 m (1A)
17. (a) (i) Take the forward direction as positive.
By v2 − u2 = 2as, we have (c) (i) The motor uses more energy because
802 – 0 = 2a(100) (1M) there is energy loss due to the friction
∴ a = 32 m s−2 of the moving parts of the motor. (2A)
When the hammer is moving up, the
So the required acceleration is
rope and the other moving parts also
32 m s−2 (forwards). (1A)
gain KE, and so the motor needs to
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.8

provide more energy for such motions. depends on the amount of gravitational
(2A)
PE converted, which in turn depends
(ii) Any ONE of the following: (1A) on her vertical position.
• Release the hammer from a higher
(ii) By KE = mv2, we have
position.
• Use a more massive hammer.
v=
19. (a) By conservation of energy, we have
mgh = mv2 (1M)
= (1M)

= 7.924 ≈ 7.92 m s−1

∴ v= = (1M)
At that instant, her vertical speed is
= 31.32 ≈ 30 m s−1 (1A) 7.92 m s−1. (1A)
Hence, the speed of the car at Q is
(c)
approximately 30 m s−1.
(b) (i) The gravitational PE loss of the car
equals its KE gain at Q and the work
done on the road by the braking force.
Mathematically,
mgh = mv2 + f s (1M) Correct diagram: 1A

(750)(9.81)(50) = (750)(272) + 80f (d) Take the upward direction as positive.

By v = u + at, we have
(1M)
7.924 = 0 + a(0.26) (1M)
∴ f = 1181 ≈ 1180 N
∴ a ≈ 30.5 m s−2
If the speed of the car has to be limited
to 27 m s−1, the required average So the average acceleration of the gymnast is
braking force would be 1180 N. (1A) 30.5 m s−2 (upwards). (1A)
(e) If the gymnast tries to reach a higher
(ii) Any ONE of the following: (1A)
• Work has to be done against friction. position, she must do more work by exerting
a greater force on the trampoline.
• Part of the mechanical energy is
When she makes contact with the
transformed into internal energy and
trampoline, her gravitational PE and the
sound energy.
work she does to the trampoline are
• The car may become more massive if
converted into the elastic PE of the
more passengers take the ride. trampoline.
(iii) Since the total mass of the car Nonetheless, some of the energy may be lost
increases, the car has more KE even as internal energy and sound energy due to
though it moves at the same speed. (1A) friction.
More work has to be done by the When the trampoline lifts her up, the elastic
braking system to take the KE away PE of the trampoline is converted into the KE
from the car and stop it. (1A) of the gymnast.
In other words, the system has to apply After she leaves the trampoline, the KE of the
a greater braking force on the car. (1A) gymnast is converted gradually into her
Quality of written communication: 1A gravitational PE , and all of the KE is
20. (a) The increase in PE when the gymnast converted when she reaches the highest
ascends is point.
PE = mgh  Candidate must state that:
= (55)(9.81)(4.2) (1M) • elastic PE is converted into KE.
= 2266 ≈ 2270 J (1A) • KE is converted into gravitational PE.
(b) (i) Her KE at the instant she just touches • the gymnast must do work/consume
the trampoline is chemical energy to reach higher.
KE = 2266 × On the other hand, candidate must:
= 1727 ≈ 1730 J (1A) • use appropriate writing style and accurate
technical words.
 By conservation of energy, the
amount of KE gained by the gymnast
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.9

• organize his answer clearly, logically and Refer to the Figure below.
coherently.
For those who can fulfill the above
requirements will get 5 to 6 marks.
21. (a) The longer pendulum has the greater
energy. (1A)
The height of a pendulum bob is given by
h = L(1 – cos θ)
where L is the length of the rod. The longer
the rod is, the higher the pendulum bob
reaches and the greater the energy it
possesses. (1A)
(b) In both cases, the gravitational PE of the
pendulum bob is converted into its KE. For
the pendulum of length ℓ, we have
mgℓ(1 – cos θ) = mvS2
• Correct axes and labels: 1A
For the pendulum of length 2ℓ , we have
• Correct scale: 1A
2mgℓ(1 – cos θ) = mvL2 (1A)
• Correct data points: 1A
Combining the two equations, we have • Correct best-fit line: 1A

=2 ⇒ = (1A) (b) (i) By conservation of energy, the

gravitational PE loss equals the KE gain,
(c) The starting height of the longer pendulum i.e.
has to be the same as that of the shorter one = (h – h0)
in order to for the two pendulums to reach
∴v2 = 2g(h − h0) (1A)
the same speed at the bottom of the swing.
(1A) (ii) From (b)(i),
Consider the Figure below. v2 = 2g(h − h0) ⇒ v2 = 2gh − 2gh0
So the slope of the graph is 2g. (1M)
From the graph, we have
2g =
= 19.6875
∴ g ≈ 9.84 m s−2
The value of gravitational acceleration
is 9.84 m s−2. (1A)

cos θ = = ⇒ θ = 60° (1A) (c) To reduce the effect of air resistance on the
ball. (1A)
22. (a)
23. (a) (i) The mechanical power delivered by the
h/m v/m s−1 v2 / m2 s−2 motor is
0.1 0.949 0.901 P = Fv = (8000)(2) = 16000 W (1M+1A)
0.3 2.214 4.902 (ii) Power loss = 20000 – 16000 = 4000 W
(1A)
0.5 2.966 8.797
0.7 3.578 12.80 (b) (i) Total mechanical power output of the
motor
= useful power to lift up the lift car +
power loss
= (8000 − 7000)(2) + 4000 = 6000 W.
(1M+1A)

(ii) Less output power is required to lift

the lift car. (1A)
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.10

(iii) The claim is incorrect. The lift will fall. ) flows downwards for a vertical distance of .
(1A)

If a drum with frictionless surface is So the PE loss is

used, the cable cannot be fixed on the EP = mgh = (g) =
drum. (1A)
24. Stick the paper strip on BC. (1A)
2. D
Take the upward direction as positive. Let v0 be the
Put the toy skier on an certain position along AB,
initial speed of the stone.
and measure its height h above the bench. Release
By v = u + at, we have
the toy skier and measure the stopping distance s
from B when it stops. (1A) 0 = v0 – gT ⇒ T=

Repeat the experiment by setting different initial At time t = , the speed of the stone is
heights and measuring the corresponding
stopping distance. (1A) v = v0 – g = v0 – g =
Plot a graph of h against s. (1A)
The KE loss of the stone equals the gravitational
By conservation of energy, PE gain, and hence
mgh = F s ⇒ h ∝ s
mgh1 = mv02 – m( )2
where F is the friction acting on the strip.
Therefore, a straight line passing through the = × mv02
origin should be obtained. (1A)
=
25. (a) From t = 0 to 5 s, Bolt accelerates from rest.
(1A)
When KE of the stone is decreased by half, it
After t = 5 s, Bolt runs at a constant speed of
reaches the height h2. Therefore half of its KE is
12 m s−1. (1A)
converted to its PE, so
(b) During the race, the chemical energy of Bolt’s
mgh2 =
body is converted into his KE and the work
done against air resistance. (2A)
Therefore, h1 : h2 = : = 3 : 2.
(c) Let f be the average resistive force acting on
Bolt during the race. By the results in (b), 3. B
(1M) When the c.g. of the block is at the right hand side
f s + 0.078E = E of X, it will topple without any energy input. So,
f (100) + (0.078)(81.58 × 103) = 81.58 × 103 work is done to rotate the block until its c.g. is just
(1M) vertically above X as shown.
∴ f = 752.2 ≈ 752 N
The average resistive force acting on Bolt
during the race is 752 N. (1A)
(d) The final speed of Bolt remains unchanged at
12 m s−1. His new average acceleration is
a= = ≈ 1.28 m s−2 (1M+1A)
The vertical distance that the c.g. raises is 2 cm.
Therefore, the minimum work done required is
Shoot-the-stars Questions (p.233) 2W.

1. A 4. (a) The volume of the asteroid is

V= πr3

= π(500 × 103)3

= 5.236 × 1017 m3
The mass of the asteroid is
M = ρV
= (2500)(5.236 × 1017)
≈ 1.31 × 1021 kg (1A)
(b) Assuming that all the energy released by the
When the tap is opened, half of the water (of mass nuclear bomb is converted into the KE of the
asteroid fragments. By conservation of
Active Physics Full Solutions to Textbook Exercises 08 Work, Energy and Power | p.11

energy, we have

KE = = 1.2 × 1018 J (1M)

The speed v of each segment is

KE = mv2

1.2 × 1018 = vy2 (1M)

∴ vy = 0.06056 ≈ 0.0606 m s−1

Therefore the maximum speed of each
segment in the y–direction is 0.0606 m s−1.
(1A)

(c) The displacement in the y–direction is

y = vyt = (0.06056)(3 × 60 × 60)
= 654.0 m ≪ 6.37 × 106 m (1M)
Yes, the Earth will be hit. (1A)