2 Force and Motion Chapter 6 Work, Energy and Power
6 Work, Energy and Power
New Senior Secondary Physics at Work (Second Edition) 1
Ó Oxford University Press 2015
2 Force and Motion Chapter 6 Work, Energy and Power
Practice 6.1 (p. 213)
1 C
2 B
3 B
4 A
Work done = Fs cos q
= 1.2 ´ 0.5 cos (90° - 40°)
= 0.386 J
5 Take the direction to the right as positive.
Net force acting on the block = 8 - 4 = 4 N
Total work done = work done by net force
= Fs
= 4 ´ 3 = 12 J
6 The upward force F has the same magnitude as the weight of bucket
W = Fs
500 = mgs
= 9 ´ 9.81s
s = 5.66 m
The distance travelled by the bucket is 5.66 m.
7 (a) Work done = PE gain by box
= mgh
= 10 ´ 9.81 ´ 0.8 = 78.5 J
(b) (i) Zero
(ii) Yes
8 Work done = Fs cos q
= 15 ´ 1.6 cos 50° = 15.4 J
9 (a) /(b) Friction = 12 cos 40° = 9.19 N
(c) Net force acting on the pram = 0
\ Total work done = 0
(d) Energy gained = total work done = 0
Practice 6.2 (p.219)
1 B
PE gained
= mgh = 50 ´ 9.81 ´ 150 = 73 600 J
2 C
mPvP 2 =mTvT 2
=
=
3 A
4 B
5 Energy stored
= work done by average force
= Fs
= 12 ´ 0.08
= 0.96 J
6 KE =mv2
5 ´ 10-4 =(80 ´ 10-6)v2
v = 3.54 m s-1
The speed of the fly is 3.54 m s-1.
7 Gain in KE = work done on the box
= Fs
= (20 cos 30° - 10)4
= 29.3 J
8 (a) PE = mgh
15 000 = 75 ´ 9.81h
h = 20.4 m
The cliff is 20.4 m high.
(b) PE = mgh
= 75 ´ 9.81 ´ (-4)
= -2940 J
(c) Greatest change = -2940 - 15 000
= -17 900 J
(d) Remain unchanged
9 (a) PE = mgh = mgd µ d
(b) PE = mgh
= mg (ut +at 2)
= mg(-gt 2)
= -mg2t 2 µ -mt 2
Practice 6.3 (p.232)
1 D
2 B
Work done against friction
= change in KE
fs =m (u2 - v2)
s =m (u2 - v2)
=´ 1500
= 25 m
3 B
Gain in KE = loss in PE = mgs
4 C
Loss in PE = gain in internal energy
mgh = mcDT
DT =
== 0.234 °C
T = 12 + 0.234 » 12.2 °C
5 (a) Max gain in PE
= mgh
= 65 ´ 9.81 ´ (5.06 - 0.9)
= 2650 J
(b) Kinetic energy, gravitational potential energy, elastic potential energy (internal energy of Isinbayeva, pole, air and mat)
6 (a) Gain in KE = loss in EPE
mv2 = 300
Þ v === 20 m s-1
The speed of the arrow is 20 m s-1.
(b) Gain in GPE = loss in EPE
mgh = 300
h =
=
= 20.4 m
The maximum height is 20.4 m.
7 Its speed will be lower than v. This is because work is done against air resistance, so the energy of the stone decreases.
8 (a) Work done = gain in KE
=mv2
=´ 0.1 ´ 52 = 1.25 J
(b) (i) By s =(u + v)t,
distance travelled
=(0 + 5)(5 ´ 10-3)
= 0.0125 m
(ii) Work done on puck = Fs = 1.25 J
\ Average force == 100 N
9 (a) When the ball bearing falls from W to X,
loss in PE = gain in KE
mgh =mv2
h =
=
= 0.204 m
Height of W above oil = 0.1 + 0.204
= 0.304 m
= 30.4 cm
(b) Loss in PE = gain in KE + work done
against friction
mgh =mv2 + fs
0.05 ´ 9.81(0.304 + 0.15)
=´ 0.05(0.5)2 + f ´ 0.15
f = 1.44 N
The average resisting force is 1.44 N.
10 The sand and small stones provides large friction on the vehicle. Part of the kinetic energy of the vehicle will become internal energy by doing work against the friction.
Also, the lane is inclined upwards, so part of the kinetic energy of the vehicle will become gravitational potential energy.
These two effects stop the vehicle using this lane.
Practice 6.4 (p.239)
1 D
Since the block moves at constant speed,
pulling force by motor = friction = F
P = Fv
F =
=
= 500 N
2 A
P = Fv
v =
=
= 0.0470 m s-1
3 D
Let F be the force provided by the engine and f be the resistive force.
Since the car accelerates uniformly, the net force acting on it is constant.
Þ F - f is constant
Þ F is constant
By v2 = u2 + 2as,
v ==
P = Fv
= F
µ
4 Average power of sports car
=
=
=
= 104 000 W
Average power of cheetah
=
=
= 9000 W
The sports car has a larger average power.
5 (a) Average Power output
=
=
=
= 20 600 W (= 20.6 kW)
(b) Work done by engine + loss in PE of counter weight = gain in PE of lift cage
Pt + Mgh = mgh
P ´ 20 + 500 ´ 9.81 ´ 60
= 700 ´ 9.81 ´ 60
P = 5890 W (= 5.89 kW)
Average power output is 5.89 kW.
6 (a) Loss in PE = mgh
= 4000 ´ 9.81 ´ 500
= 1.96 ´ 107 J
(b) Power output =
=
= 1.96 ´ 107 W
(c) The water is still moving after passing through the turbine, i.e. the water does not transfer all its kinetic energy to the turbine.
Some energy of the water is converted into the internal energy of the water, turbine and pipe.
Revision exercise 6
Concept traps (p.242)
1 F
A force does no work if it is perpendicular to the displacement.
2 T
3 F
Consider a car moving down an inclined road at a constant velocity. Its PE decreases but its KE remains unchanged.
4 F
Consider an object on the floor of a train which accelerates forwards. The friction acting on the object points forwards. The work it does becomes the KE of the object.
Multiple-choice questions (p.242)
5 D
6 B
Gain in PE = mgh
= 50 ´ 9.81 ´ 0.005
= 2.45 J
7 C
Work done
= Fs = 20 ´ 2 ´ p ´ 0.3 = 37.7 J
8 C
9 C
Apply loss in PE = gain in KE.
When the box is released from X,
mgh =mv2 (1)
When the box is released from Y,
mgH =m(2v)2 = 2mv2 (2)
(2) ¸ (1),
= 4
H = 4h
10 B
Work done by Stephen
= KE gained by ball
=mv2
=´ 0.1 ´ 52
= 1.25 J
\ (1) is correct.
Gain in PE = loss in KE
mgh =mv2
h =
\ (2) is correct.
When the ball moves down the rail, the work done by its weight becomes its KE.
\ (3) is incorrect.
11 A
Sum of KE + PE = initial PE
= mgh
= 0.5 ´ 9.81 ´ 0.1
= 0.491 J
\ (1) is correct.
The speed of the bob at Y is independent of the mass of the bob.
\ (2) is incorrect.
By conservation of energy, the maximum potential energy of the bob on the other side is the same as that at X, i.e. the bob will move to the same level as X.
\ (3) is incorrect.
12 A
Loss in PE = gain in KE
mgh =mv2
Þ h === 0.815 m
\ (1) is correct.
mgh =mv2
v =
=
= 2.49 m s-1
\ (2) is incorrect.
Since the track is smooth, no work is done against friction. Z is at the same level as X, so the ball has the same KE and speed at these two points.
\ (3) is incorrect.
13 A
Let f be the braking force.
Loss in KE = work done against braking force
mv2 = fs
s =
=
=
=
=
14 C
Loss in KE = gain in PE + work done against
friction
KEi - KE = mgh + fs
= mg ´+ fs
=s
KE = KEi -s
KE varies linearly with s, and its value is maximum when s = 0.
PE = mgh = mg ´µ s
Also,
gain in PE
= loss in KE - work done against f
< loss in KE
PE depends linearly on s, and its value is smaller than the loss in KE.
15 A
The displacement of R is 0.
16 (HKCEE 2007 Paper 2 Q31)
17 (HKCEE 2008 Paper 2 Q28)
18 (HKCEE 2009 Paper 2 Q6)
19 (HKDSE Practice Paper 2012 Paper 1A Q11)
20 (HKDSE 2012 Paper 1A Q9)
21 (HKDSE 2013 Paper 1A Q12)
22 (HKDSE 2014 Paper 1A Q6)
Conventional questions (p.245)
23 (a) KE =mv2 1M
=´ 0.057 ´
= 106 J 1A
(b) Work done by the racket
= KE gained by the ball
Fs = 106 1M
F =
=
= 1060 N 1A
The average force is 1060 N.
24 (a) /(Each correct force) 2 ´ 1A
(b) (i) By v2 = u2 + 2as, 1M
1.22 = 0 + 2a ´ 0.8
a = 0.9 m s-2 1A
The man’s acceleration is 0.9 m s-2.
(ii) Work done on the box
= KE gain by the box
=mv2 1M
=´ 5 ´ 1.22 = 3.6 J 1A
(iii) Friction f does the work. 1A
Þ fs = 3.6 1M
Þ f === 4.5 N 1A
25 (a) Loss in PE = gain in KE
mgh =mv2 1M
9.81 ´ (100 - h) =´ 252
h = 68.1 m 1A
Position B is 68.1 m above sea level.
(b) Loss in PE = gain in KE + work done
against friction
mgh =mv2 + fs 1M + 1M
300 ´ 9.81 ´ (100 - 36)
=´ 300 ´ 252 + f ´ 500
f = 189 N 1A
The average friction is 189 N.
(c) Any three of the following: 3 ´ 1A
Gravitational potential energy, kinetic energy, internal energy, sound energy
26 (a) (i) Initial KE =mv2 1M
=´ 1 ´ 42 = 8 J 1A
(ii) Loss in KE = gain in PE + work
done against friction
8 = mgh + fh 1M + 1M
h =
=
= 0.540 m 1A
The metal cylinder can reach a high of 0.540 m.
(b) mv2 = mgh + fs
´ 1 ´ v2 = 1 ´ 9.81 ´ 3 + 5 ´ 3
v = 9.43 m s-1 1A
The minimum speed is 9.43 m s-1.
(c) Any two of the following: 2 ´ 1A
Use a heavier metal cylinder.
Put the bell higher.
Increase the friction between the cylinder and the board.
27 (a) Work done = Fs 1M
= (mg sin q + f ) s
= (30 ´ 9.81 sin 20° + 80) 3
= 542 J 1A
(b) Total gain in KE and PE
= gain in PE
= mgh 1M
= 30 ´ 9.81 ´ 3 sin 20° = 302 J 1A
(c) No, 1A
this is because part of the work done on the box becomes internal energy in doing work against friction. 1A
(d) A smaller force is needed in pushing the box up the inclined plane. 1A
28 (a) The ball should be released from 1 m above the ground. 1A
(b) (i) Let H be the height from which the ball is released.
Loss in PE = work done against
friction
mg(H - h) = mgH ´ 10% 1M
H - 1 = 0.1H
H = 1.11 m 1A
The ball should be released from 1.11 m above the ground.
(ii) It becomes the internal energy of the ball and the rail 1A
(c) No, 1A
this is because extra energy is given to the ball when she pushes it. 1A
29 (a) (i) Distance travelled
=(u + v) t 1M
=(10 + 25) 2
= 35 m 1A
(ii) Average power output
=
= 1M
=
= 55 100 W (= 55.1 kW) 1A
(iii) No. 1A
Since F = ma + f,
a and f being constant
Þ F being constant 1A
By P = Fv, v varying and F being constant Þ P varying 1A
(b) Resultant force down the slope
= mg sin q + f
= 300 ´ 9.81 sin 5° + 1600
= 1856.5 N 1M
When the motorcycle travels uniformly at the maximum speed, the forward force produced by the engine is 1856.5 N.
By P = Fv,
maximum speed =
=