Jurong Junior College
2014 JC1 H1 and H2 Physics (8866 and 9646)
Tutorial Topic: 5. Work, Energy and Power
Learning Outcomes (Candidates should be able to:)
(a) show an understanding of the concept of work in terms of the product of
a force and displacement in the direction of the force.
(b) calculate the work done in a number of situations, including the work
done by a gas which is expanding against a constant external pressure:
W = pV.
(c) give examples of energy in different forms, its conversion and
conservation, and apply the principle of energy conservation to simple
examples.
(d) derive, from the equations of motion, the formula Ek = ½ m v2.
(e)
(f)
(g)
(h)
(i)
(j)
(k)
Questions
1
2
3, 7, 9, 10, 11, .
11, refer to
lecture notes.
recall and apply the formula Ek = ½ m v2.
3, 11, .
distinguish between gravitational potential energy, electric potential Refer to lecture
energy and elastic potential energy.
notes.
show an understanding of and use the relationship between force and 4
potential energy in a uniform field to solve problems.
derive, from the defining equation W = Fs, the formula Ep = mgh for 11, refer to
potential energy changes near the Earth’s surface.
lecture notes.
recall and use the formula Ep = mgh for potential energy changes near 5, 8, 11, .
the Earth’s surface.
show an appreciation for the implications of energy losses in practical 5
devices and use the concept of efficiency to solve problems.
define power as work done per unit time and derive power as the product 5, 6, refer to
of force and velocity.
lecture notes.
Section A: Fundamental/conceptual questions
F
1.
θ
s
A object is pulled by force F over a horizontal distance s. Calculate the work done if
(i)
(ii)
(iii)
(iv)
F is 1 N, θ is 0 and s is 2 m.
[W = F s cos θ]
(1)(2) cos 0o = 2 J
[2 J]
F is 4 N, θ is 30 and s is 5 m.
[W = F s cos θ]
(4)(5) cos 30o = 17.3 J
[17.3 J]
F is 7 N, θ is 60 and s is 8 m.
[W = F s cos θ]
(7)(8) cos 60o = 28 J
[28 J]
F is 10 N, θ is 90 and s is 11 m.
[W = F s cos θ]
(10)(11) cos 90o = 0 J
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[0 J]
2.
A gas expands by 100 cm3 against a constant pressure of 1.25 × 105 Pa. Calculate the
work done by the gas.
[12.5 J]
[W = pV]
(1.25 × 105)(100)(10-2)3 = 12.5 J
3.
On braking, 500 kJ of heat was produced when a vehicle of total mass 1600 kg was
brought to rest on a level road. What was the speed of the vehicle just before the brakes
were applied?
[25 m s-1]
Loss in KE of the vehicle = heat dissipated
1
2
(1600)v 2  500  103
v = 25 m s-1
4.
5.
A charged body placed in a uniform electric field experiences an electric force of 10 N.
(i)
What is the change in the electric potential energy of the body if it moves a
displacement of 5.0 cm in the direction of the electric force?
[Fx = U]
[0.5 J]
U = Fx = (10)(0.05) = 0.5 J
(ii)
Is there an increase or a decrease in the electric potential energy of the charged
body?
U is negative means there is a decrease in the electric potential energy of the
charged body
A crane lifts a load of 8000 N at constant speed through a vertical distance of 20 m in a
time of 4 s.
(i)
(ii)
What is the average useful power during this operation?
Average useful power
total useful work done 8000  20


 40 kW
total time taken
4
If the efficiency of the motor is 20%, what is the electrical power supplied to the
motor?
[200 kW]
useful output energy
useful output power
Efficiency 
x100% 
x100%
total input energy
total input power
40  103
 100%
total input power
Total input power = 200 kW
20% 
6.
[40 kW]
TYS Page 67 Question 9 [N09/I/12]
At constant velocity, resultant force on boat is zero, hence
applied force from boat engine = total resistive force experienced by boat
= kv2
power required from boat engine
= applied force from boat engine × velocity of boat = (kv2)(v) = kv3
72000 = kv3 = k(123)
eqn 1
Page 2
With only 1 engine, power = 36000 W, hence
36000 = kv3
eqn 2
1
Dividing eqn 1 by eqn 2 gives v3 = 2(123)
So, v = 9.5 ms-1
Section B: Short examination-style questions
7.
The diagram on the right
shows an arrangement used to
find the output power of an
electric motor. The wheel
attached to the motor's axle
has a circumference of 0.5 m
and the belt which passes over
it is stationary when the
weights have the values
shown.
If the wheel makes 20
revolutions per second, what is
the output power?
[300 W]
P
8.
Fs
(20)(0.5)
s
 F     50  20 
 300 W
t
1
t 
The diagram shows two bodies X and Y
connected by a light cord passing over a
light, free-running pulley. X starts from rest
and moves on a smooth plane inclined at
30 to the horizontal.
Calculate the total kinetic energy of the
system when X has travelled 2.0 m along
the plane.
[59 J]
Using conservation of energy,
KE gain by X + KE gain by Y + GPE gain by X = GPE loss by Y
Total KE of system + (4.0)(9.81)(2.0 sin 30o) = (5.0)(9.81)(2.0)
Total KE of system = 59 J
9.
Discuss the energy changes which take place for a system comprising a mass
suspended from the end of a spring and then made to oscillate vertically. You need to
refer to the different forms of energy involved.
(Assume the oscillation is small enough so that the spring is never compressed but
always stretched by different extent during the oscillation.) Assume the oscillation
is started by displacing the mass vertically downwards from its equilibrium position
thus stretching the spring and giving it elastic potential energy (EPE). When the
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mass is released and the oscillation begins, the EPE stored in the spring is
converted into kinetic energy (KE) and gravitational potential energy (GPE) of the
mass as it moves upwards. The KE of the mass is maximum when it moves at
maximum speed past the equilibrium position. After the mass passes the
equilibrium position, its KE and the EPE stored in the spring are converted into
GPE as it continues to move upwards. When the mass reach the highest point in
the oscillation, its KE is zero and its GPE is maximum whereas the EPE stored in
the spring is minimum. As the mass move from the highest point to the equilibrium
position, its GPE is converted to its KE and EPE stored in the spring. As it moves
from the equilibrium position to the lowest point, its KE and GPE are converted to
EPE stored in the spring, thus completing one oscillation. Dissipative forces will
gradually convert the mechanical energy into thermal energy so the amplitude of
the oscillation will decrease with time and eventually become zero.
Section C: Long examination-style questions
10.
(a)
Starting with the definition of work, deduce the change in the gravitational potential
energy of a mass m, when moved a distance h upwards against a gravitational
field of field 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 Ek = ½mv2.
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(c)
A cyclist, together with his bicycle, has a total mass of 90 kg and is travelling with a
constant speed of 15 m s1 on a flat road at A, as shown below.
He then goes down a small slope to B, descending 4.0 m in doing so. Calculate
(i)
the kinetic energy at A,
½mv2 = ½(90)(15)2 = 10.1 kJ
[10.1 kJ]
(ii) the loss of potential energy between A and B,
mgh = (90)(9.81)(4.0) = 3.53 kJ
[3.53 kJ]
(iii)
the speed at B, assuming that all the lost potential energy is transformed into
kinetic energy of the cyclist and bicycle.
Loss in GPE = gain in KE
[17.4 m s1]
2
2
(90)(9.81)(4.0) = ½(90)v  ½(90)(15)
v = 17.4 m s-1
(d)
(i)
A cyclist traveling at a constant speed of 15 m s1 on a level road provides a
power of 240 W. Calculate the total resistive force.
For constant speed motion,
[16 N]
force provided by cyclist = total resistive force F.
Power from cyclist = (Force provided by cyclist)(velocity of cyclist)
240 = (F)(15)
F = 16 N
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(ii)
(c)
The cyclist now travels at a higher constant speed. Explain why the
cyclist needs to provide a greater power.
1. Higher speed also means larger value for total resistive force F since
air resistance increases with speed.
2. Since the cyclist is moving at constant speed, force by the cyclist is
equal to total resistive force, force by cyclist increase.
3. The power required of the cyclist is the product of F and v so it will be
higher.
It is often stated that many forms of transport transform chemical energy into
kinetic energy. Explain why the cyclist travelling at constant speed is not making
this transformation. Explain what transformations of energy are taking place.
Since the cyclist is travelling at constant speed, there is no increase in kinetic
energy. The chemical energy from the cyclist is being converted to thermal energy
in the gears, tyres, road and air when overcoming resistive forces. Some of the
chemical energy is also converted to sound energy.
11.
(a)
The minimum flying speed for a bird called house-martin is 9.0 m s1. It reaches
this speed by falling from its nest before swooping away. Calculate the minimum
distance its nest must be above the ground.
[4.13 m]
Loss in GPE = gain in KE
m(9.81)h = ½m(9.0)2
h = 4.13 m
(b)
A house-martin has a mass of 120 g. When it returns to its nest, it is travelling
horizontally at P with a speed of 13.0 m s1 and at a distance 7.5 m below its nest.
It then glides upwards to the nest, as shown in the figure below.
Neglecting any air resistance, calculate
(i)
(ii)
the kinetic energy of the house-martin at P,
KE at P = ½(120 x 10-3)(13.0)2 = 10.1 J
[10.1 J]
the gain in potential energy as it glides upwards to its nest,
gain in GPE = (120 x 10-3)(9.81)(7.5) = 8.83 J
[8.83 J]
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(iii)
(iv)
(c)
its kinetic energy as it reaches its nest,
Gain in GPE = loss in KE
8.83 = 10.1  KE as it reaches its nest
KE as it reaches its nest = 1.27 J
its speed as it reaches its nest,
½(120 x 10-3)v2 = 1.27
v = 4.60 ms-1
[1.27 J]
[4.60 m s-1]
Discuss qualitatively how air resistance affects, if at all, each of your answers in
(b).
(b)(i) No effect.
(b)(ii) No effect (but the house-martin may not reach its nest if its KE at P is less
than the sum of GPE required to gain that height and the energy required to
overcome air resistance).
(b)(iii) Decreased (since there is work done to overcome air resistance).
(b)(iv) Decreased (since it reaches its nest with less KE).
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