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Innova Junior College
JC 1 Physics
Tutorial 5 Work, Energy & Power
Work
1. (a) Define work.
(b) A 50 kg box is pulled 40 m along a horizontal floor by a constant force exerted by a
person, P = 100 N which acts at a 37°. The floor is rough and exerts a frictional force
f = 50 N.
(i) Determine the work done on the box by the individual force acting on it.
[0 J, 0 J, 3190 J, - 2000 J]
(ii) Hence or otherwise, calculate the net work done on the box.
[1200 J]
(a) Work is the product of the force, F and the displacement, s in the direction of the
force.
(b) The four forces acting on the box are force P, frictional force f, weight W and normal
force N.
(i) WW = mgx cos 90° = 0
WN = Nx cos 90° = 0
WP = 100 × 40 × cos 37° = + 3190 J
Wf = 50 × 40 × cos 180° = -2000 J
(ii) Net work done on box, Wnet = Ww + WN + WP + Wf
= 0 + 0 + 3190 + ( - 2000)
= 1200 J
Alternatively, Wnet = Fnets cos θ
Since the box is moving along the floor, the net force acting on the box will be
acting horizontally, i.e. its weight and normal force does not have any component
in the computing of net force.
Fnet = P cos 37° - f
= 100 cos 37° - 50
= 29.9 N
Wnet = 29.9 × 40 cos 0°
= + 1200 J
2.
F/N
400
200
0
- 200
x/m
2.0
4.0
6.0
8.0 10.0 12.0 14.0 16.0
- 400
The force of a force acting on an object varies as shown. Determine the work done on
the object by this force to move the object:
(a) from x = 0 to x = 10 m and,
(b) from x = 0 to x = 15 m
[2800 J, 2100 J]
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(a) from x = 0 to x = 10 m:
W = Area of the trapezium ABCD
 ½  4.0+10.0  400 
= + 2800 J
(b) from x = 0 to x = 15 m:
W = Area of the trapezium ABCD + Area of the trapezium DEFG
 ½  4.0+10.0  400   ½  2.0+5.0  200 
= + 2100 J
Energy
3. (a) For a swinging pendulum, describe the various energy transformations and how the
Law of Conservation of Energy can be applied.
(b) A pendulum bob of mass 0.20 kg is suspended by means of a string of length 2.0 m
from a fixed point. The bob is pulled to the side with the string taut until the string
makes an angle 30 with the vertical.
(i) the gain of potential energy when the bob is moved aside and,
[0.53 J]
(ii) the speed as the bob (after being released) passes through the position where
the string is vertical.
[2.3 m s-1]
(a) With reference to the diagram. At B, the gravitational potential
energy (Ep) is taken to be zero (as the reference level).
At A and C, the pendulum comes to rest momentarily and its
speed is zero. Thus it has zero kinetic energy (Ek). It is also at
the maximum height, thus its Ep is at its maximum value.
From the Law of Conservation of Energy, the sum of Ep and Ek
must be constant, assuming air resistance and friction are
negligible.
Hence, as the pendulum moves from A to B, it loses Ep and
gains Ek. As it moves from B to C, it gains Ep and loses Ek.
And the cycles continues as the pendulum swings back from C
to B to A.
A
C
B
(b) (i) Take Ep to be zero at the lowest point of the pendulum.
Height gained = 2.0 – 2.0 cos 30 = 0.268 m
Gain in Ep = mgΔh = 0.20 × 9.81 × 0.268 = 0.53 J
(ii) From the Law of Conservation of Energy,
(Ep)A + (Ek)A = (Ep)B + (Ek)B
0.526 + 0 = 0 + ½ mv2
0.526 = ½ (0.20) v2
v = 2.3 m s-1
4. J02/I/18
A trolley runs from P to Q along a track. At Q its potential energy is 50 kJ less than at P.
trolley
P
Q
At P, the kinetic energy of the trolley is 5 kJ. Between P and Q, the work the trolley does
against friction is 10 kJ. Determine the kinetic energy of the trolley at Q.
[45 kJ]
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Using Law of Conservation of Energy,
Ek P  EP P  W  Ek Q  EP Q
5  0   10   Ek Q   50 
(Ek)Q = 45 kJ
5. A 4.5 kg block is thrust up a 30o inclined rough plane with an initial speed of 5.0 ms-1.
(a) Determine the maximum height that the block travels up the plane, given that the
friction acting on the block is 15 N.
[0.76 m]
(b) Hence determine the speed of the block when it returns to the bottom of the inclined
plane.
[2.2 m s-1]
(a) Using Law of Conservation of Energy,
 Ek A   EP A  W   Ek B   EP B
½mv 2  0  fs cos  0  mgh
h
½  4.5  5.02  15 
cos180  4.5  9.81h
sin30
h = 0.759
h = 0.76 m
B
s
A
(b)  Ek B   EP B  W   Ek  A   EP  A
h
30°
0  mgh  fs cos  ½mv 2  0
0.759
4.5  9.81 0.759  15 
cos180  ½  4.5v 2
sin30
v = 2.2 ms-1.
6. A 2.0 kg block is released from point A. It travels down the track, hits a spring of force
constant k = 225 N m-1 at D and compresses the spring 0.20 m from its equilibrium
position before coming to rest momentarily. The track is frictionless except the rough
surface between B and C, which is of length 2.0 m.
(a)
(b)
(c)
(d)
(e)
Calculate the speed of the block at B.
[4.4 m s-1]
Calculate the energy stored in the spring after it is compressed by the block. [ 4.5 J]
Deduce the energy lost to friction.
[15J]
Find the average frictional force acting against the block.
[7.6 N]
If the block rebound and moved towards C after the spring uncompressed itself, find
the position along BC at which the block will finally come to rest, taking C as the
reference. State assumptions that you make in your calculations.
[0.60 m from C]
(a) Using Law of Conservation of Energy,
 Ek A   EP A  W   Ek B   EP B
(Take (Ep)B to be zero)
0  mgh  0  ½mv 2  0
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2.0  9.81 1.0  ½  2.0v 2
v = 4.43
v = 4.4 m s-1
(b) E.P.E stored = ½ kx2
= ½ (225) (0.20)2
= 4.5 J
(c) Consider the instant when the block compresses the spring and stops momentarily,
Using Law of Conservation of Energy,
 Ek A   EP A  W   Ek D   EP D
0  mgh  W  0  0  E.P.E
2.0  9.81 1.0  W  4.5
Work done by friction = - 15.1 J
Work done against friction = 15 J
(d) Work done against friction = frictional force  displacement
15.12 = f  2.0
f = 7.56
f = 7.6 N
(e)  Ek D   EP D  W   Ek BC   EP BC
0  0  E.P.E  fs cos180  0  0
7.56s cos180   4.5
s = 0.595
s = 0.60 m (away from C)
Power
7. N97/I/6
A force of 1000 N is needed to lift the hook of a crane at a steady velocity. The crane is
then used to lift a load of mass 1000 kg at a velocity of 0.50 m s-1. Calculate the power
developed by the motor of the crane is used in lifting the hook and the load.
[5400 W]
At constant velocity,
Fmotor = total weight of hook and load
= 1000 + 1000 (9.81)
= 10810 N
P  Fv
= 10810 × 0.50
= 5400 W
8. The hydroelectric power station has a power output of 2.0 MW when water passes
through its turbines at a rate of 1.4 m3 s-1. The water is supplied from a reservoir which is
750 m above the power station turbines.
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(a) Calculate the mass of the water passing through the turbines each second.
[1.4  103 kg s-1]
(b) Calculate the loss of potential energy per second of the water flowing between the
reservoir and the power station turbines.
[1.0  107 W]
(c) Calculate the efficiency of the power station.
[19%]
(a)
m V  V

    1.4  1000
t
t
t
= 1.4 × 103 kg s-1
(b) loss in Ep per second =
(c)  
mg h m
 g h  1.4  103  9.81 750   1.0  107 W
t
t


useful energy ouitput
 100%
total energy input
2.0  106
 100%
1.03  107
= 19%

9. Long Question (N99/III/1)
(a) Starting with the definition of work, deduce the change in the gravitational potential
energy of a mass m, when moved a distance of h upwards against a gravitational
field of strength g.
(b) By using the equations of motion, show that the kinetic energy Ek of an object of
mass m travelling with speed v is given by
1
Ek  mv 2
2
(c) A cyclist, together with his bicycle, has a total mass of 90 kg and is travelling at a
speed of 15 m s-1 on a flat road at A. He then goes down a slope to B, so
descending 4.0 m.
Calculate
(i) the kinetic energy at A
[1.0  104 J]
(ii) the loss of potential energy between A and B.
[3.5  103 J]
(iii) the speed at B, assuming that all the lost potential energy is transformed into
kinetic energy of the cyclist and bicycle.
[17 m s-1]
-1
(iv) A cyclist travelling at a constant speed of 15 m s on a level road provides a
power of 240 W. Calculate the total resistive force.
[16 N]
(v) The cyclist now travels at a higher constant speed. Explain why the cyclist needs
to provide a greater power.
(d) It is often stated that many forms of transport transform chemical energy into kinetic
energy. Explain why a cyclist travelling at a constant speed is not making this
transformation. Explain what transformations of energy are taking place.
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(a) Refer to Lecture Notes
(b) Refer to Lecture Notes
(c) (i) Ek at A = ½ mv2 = ½ (90)(15)2 = 10 125 J = 1.0  104 J
(ii) Take Ep to be zero at the ground level,
ΔEp = 0 – mgh =0 – (90)(9.81)(4.0) = – 3532 = – 3.5  103 J
Loss of potential energy between A and B = 3.5  103 J
(iii) By Law of Conservation of Energy,
 E k  A   E P  A   E k  B   E P B
10125  3532  ½  90v 2  0
v = 17 m s-1
(iv) P = Fv
240 = Fcyclist × 15
Fcyclist = 16 N
Since acceleration is zero, net force is zero. Thus f = Fcyclist = 16 N.
(v) With P = Fv, at a higher constant speed, a greater power would be needed to
produce the same force to overcome the same resistive force.
Moreover, resistive force is proportional to speed. At a higher constant speed, the
total resistive force would be greater. Hence, a greater power is needed to
produce a greater force to overcome the greater total resistive forces.
(d) A cyclist travelling at a constant speed has constant kinetic energy. There is no work
done to increase the kinetic energy of the cyclist, hence chemical energy is not
converted into kinetic energy.
In addition, to maintain constant speed (i.e. without slowing down), the cyclist has to
overcome the resistive forces. Hence the chemical energy is transformed into heat
energy in the gears, tyres and on the road as well as sound energy.
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