Download Exercise 4 Solution

Exercise 7 (Work, energy and power) Suggested Solutions
1. (a) (i) Work done by the force = F s
= 50 N x 0.1 m = 5 J
(ii) Work done by the force = increase in KE
0.5 = ½ mv2
=> 0.5 = ½ (0.1) v2 => v = 5 m s-1
(b) F = ( mv – mu ) / t =>
[(0.1)(10) – (0.1)(-10)] / 0.2 = 10 N
(c) The first statement is false. By Newton’s first law, a body will move with constant velocity if there is no
external force acting on it. Since there is no friction, the disc can therefore move with constant velocity without
a driving force.
The second statement is true. By Newton’s second law, force is equal to rate of change of momentum.
The faster the disc moves toward the wall, the greater the change of momentum. So the average force exerted
on the wall would be greater.
2.
(c) (i) Acceleration = 8 / 5 = 1.6 m s-2
(ii) Stopping distance = area under graph
= ½ x 8 x (15-5) = 40 m
(iii) Deceleration = 8 / (15-5) = 0.8 m s-2
Frictional force = ma = 60 (0.8) = 48 N
(d) The sledge is released from the same height in both cases. The sledge acquires equal speeds (or KE) at
B in both cases. So the stopping distance along BC would remain unchanged.
3. (a) Loss in PE = mgh = 75 x 9.81 x 50 = 36.8 KJ (36788 J)
(b) Loss in PE = gain in KE + work done against resistive force
36788 = (1/2) x 75 v2 + 200 x 50 / sin 20o
v = 14.2 m/s
(c) 36788 = (1/2) x 75 v2 + 200 cos 20o x 50 / sin20o
v = 15.8 m/s
4.(a) Let f be the air resistance.
Work done by air resistance = kinetic energy loss of the car
1
f  50 =
 500  60 2
2
f = 18 000 N
(Alternative method: F = ma,
v2
60 2
f=m
= 500 
= 18 000 N)
2a
2  50
The average air resistance required is 18 000 N.
(b) (i)
Let h be the height of the plane. By the conservation of
energy,
KE = PE + work done against resistance
1

Δ mv 2  = mgh + fs
2

1
h
 500  (602  202) = 500  10  h + 1500 
2
sin 30
h = 100 m
(ii) I will use parachute to stop the car.
It is because from (i) such inclined plane should be 100 m
high and 173 m across, where 100 m is roughly the height of
a 28-storey building. It is not practical to use this method.
In addition, such inclined plane can only slow down the car,
but not stopping the car.
5. (a) Work done = Fs
= 50 x 2.5 × 10–2 = 1.25 J
(b) Work done = increase in KE
1.25 = (1/2) (7.5 × 10–3) v2
v = 18.3 m/s
(c) v2 = u2 + 2as
0 = 18.32 + 2 (-9.81) s
s = 17.0 m
MC
1-5
AC E DA
6-10
C AA B C
11-15
BB DBA
Explanations to MC
1.
Loss in PE = gain in KE
mgx = m v2 / 2
x = v2 / 2g = 2.62 / 2 (10) = 0.338 m
2.
Work done = increase in KE
= (1/2)(5) (102 - 62) = 160 J
4.
(1) is wrong because both spheres have the same acceleration.
For (2), s = (1/2)gt2
For sphere A, 4h = (1/2)g tA2
For sphere B, h = (1/2) g tB2
So tA = 2tB
For (3), when reaching the ground,
KE of A = mg (4h) = 1(g) (4h) = 4gh
KE of B = mg (h) = 2 g (h) = 2 gh
KE of A = 2 KE of B
5.
mg (0.3+0.15) = mvB2 / 2 => vB = 3 m s-1
mg (0.3-0.1) = m vA2/ 2 => vA = 2 m s -1
6.
Minimum power output = mgh / t = 120 (500) (10) / 60 = 10000 W
7.
PE = mgh = mg (H-s) where H is the initial height
PE = - mgs + mgH
The graph is a straight line with negative slope -mg, and positive intercept mgH.
8.
Work done against friction = Fs
= F v t = 5 x 10 x 4 = 200 J
9.
Useful output power = KE gained / time
= [m v2 / 2] / 4 = [m (u+at)2 / 2] /4
= [2000 (3 x 4)2 / 2] /4 = 36 kW
10. For (1), PE = mgh, obviously the PE is increasing.
For (2), tension = 24 N, mg = 2 x 10 = 20 N, so net force on the block is T - mg = 24 - 20 = 4 N, therefore the block
must be accelerating, i.e. the kinetic energy must be increasing.
(3) is wrong because
average power = (PE gained + KE gained) / time
PE gained = 2 x 10 x 4 = 80 J
acceleration = F/m = (24 - 20) / 2 = 2 m s-2.
speed gained v = u + at = 0 + 2 x 2 = 4 m s-1.
So KE gained = mv2/2 = 2 x 42 / 2 = 16 J.
=> average power = (80+16)/ 2 = 48 W.
Alternative faster method :
Power developed by the machine
= Fs / t = rate of work done by tension = 24 x 4 / 2 = 48 W
11. Energy loss = KE loss
1
x 0.04 (3302 – 2082) = 1312 J
2
12. Output power = mgh / t = mgv = 1000 x 10 x 0.1 = 1000 W
13. Work done on the box by the force = Fs = F (vt) = 10 x (2 x 3) = 60 J
14. Work done on the box = Fs + mgh = 10 x 3 + 20 x 0.5 = 40 J
15. (1) is incorrect, because the net force is mg – f, where f is the air resistance. But as the object falls downward, speed
increases, so air resistance increases. Thus mg – f decreases, and hence the acceleration decreases.
(2) is correct, refer to the above explanation.
(3) is wrong, because loss in PE = gain in KE + WD against air resistance.