Work, Energy and
Power
LESSON OBJECTIVES
At the end of the lessons, pupils should be able to :
• Show understanding that kinetic energy, potential energy (chemical, gravitational, elastic), light energy, thermal energy, electrical energy and nuclear energy are examples of different forms of energy.
• State the principle of the conservation of energy and apply the principle to new situations or to solve related problems.
• State that
– kinetic energy Ek = ½ mv2 and
• Recall and apply the relationship work done = force x
distance moved in the direction of the force to new situations or to solve related problems.
• Recall and apply the relationship power = work done/time taken to new situations or to solve related problems.
• Calculate the efficiency of an energy conversion using the formula efficiency = energy converted to useful output / total energy input
Lesson Overview
• Energy
• Principle of Conservation of Energy
Work
In Physics, work is said to be done when a force produces motion.
RECALL FROM SEC 1’s WORK
Conditions For Work Done (Very Important !)
1. There is a force acting on an object. 2. The object moves.
3. The movement of the object is in the direction of the force.
object
Force applied by Mario The object moves in
QUESTION
Is there work done??
QUESTION
Is there work done??
d
F
W
in
Newton, N
in
metre, m
in
Joule, J
Work
distance moved in the direction of the force
Work done = force x
Note:
The force exerted on the object must be in the same direction as the
Moment Work
M = F x d W = F x d
d is the perpendicular distance between
applied force and pivot.
d is the distance moved
in the direction of the force.
Measured in
Newton-metre, Nm
Measured in
Joule, J
Example 1
A box of 10 kg moves 3 m after being pushed by a constant force of 4 N. What is the work done by the push?
W = F x d
= 4 x 3
= 12 J
Find the work done by gravity when a diver of mass 60 kg falls through a height of 11m.
Work done by gravity on diver
= F x d = Wdiver x d
= mdiverg x height
= (60 x 10) x 11
= 6600 J or 6.6 kJ
Energy
• Defined as the capacity to do work
• The SI unit of energy is joule (J)
• There are different form of energy namely:
Kinetic energy
Gravitational potential energy
Elastic potential energy e.g. rubber band
Chemical potential energy e.g. food, fuel, battery
When a body is in motion, it possesses kinetic energy. It can be shown that the amount of kinetic energy possessed by a moving body is given by:
K.E. = ½ m v2
Kinetic Energy (K.E.)
where m : mass of the body in kg;
v : velocity of the body in ms-1
What is the instantaneous kinetic energy of a car of mass 1200 kg
(a) moving at a velocity of 20 ms-1? (b) at rest?
(a)moving at a velocity of 20 ms-1?
K.E. = ½ mv2
= ½ (1200)(20)2
= 240000 J or 240 kJ
Instantaneous K.E. of the object is 240 kJ.
(b) at rest v = 0 ms-1
K.E. = ½ mv2
= ½ (1200)(0)2 = 0 J
Instantaneous K.E. of the object is 0 J.
Note: Object has K.E. only when in motion….
A body possesses gravitational potential energy because of its relative position to the ground. Again it can be shown the gravitational potential energy possessed by a body is given by:
where m is the mass of the body in kg;
g is gravitational field strength in Nkg-1;
h is the height above the ground in m.
2.2 Gravitational Potential Energy (P.E.)
G.P.E. = m g h
40 m
5 N 3 kg
5m
Terry pushes a box of 3 kg up a smooth slope with a
force of 5 N. If the box has moved 40 m and the height of the slope is 5 m,
(a) what is the gravitational potential energy gained by the box?
(b) how much work is done by Terry?
40 m
5 N 3 kg
5m
(a)what is the gravitational potential energy gained by the box?
P.E. = m g h
= (3) (10) (5) = 150 J
Example 4
40 m
5 N 3 kg
5m
(b) how much work is done by Terry?
W = F × d = 5 × 40 = 200 J
Example 4
Principle of Conservation of Energy
It states that:
One of the most fundamental laws in Physics
“Energy cannot be created or destroyed,
but only changes from one form to
Stored Chemical Energy
Gravitational
Potential Energy
Kinetic Energy
Work in driving the nail +
Sound + Heat
Gravitational Potential Energy
Kinetic Energy
Ground
Potential energy
decreases as kinetic
energy increases.
However, at any
moment of time, total energy remains the same.
Just before ball is
dropped, total energy of the ball is made up of P.E. only.
In mid air, total energy of the ball is made up of P.E. and K.E.
Just before hitting the ground, total energy is made up of K.E. only.
Example 7:
The diagram shows the heights above the ground of some points on the track of a roller coaster.
What is the speed of the roller coaster at P, Q and R? Assume negligible friction and air resistance.
QUESTION:
Assume negligible friction and air resistance.
Therefore we can assume there is no:
- heat energy
- sound energy
Can you tell me what are the implications of these assumptions?
All the energy is
Example 7:
A
Key point:
The total amount of energy in a system is always constant. i.e. Total Energy = K.E. + P.E.
P.E. only
K.E. only
K.E + P.E.
At point A,
The roller coaster is not moving,
** K.E. = 0 J
Total energy = G.P.E.
= m g h1
= m x 1 0 x 16.2
= 1 6 2m J
At point P,
** P.E. = 0 J
Total energy = K.E.
162m = ½ mvP2
162m = m (½ vP2)
162 = ½ vP2
vP2 = 324
vP = 18 ms-1
At point Q,
Total energy = K.E. + P.E.
162m = ½ mvQ2 + mgh
2
162m = m (½ vQ2 + 10 x 11.2)
162 = ½ vQ2 +112
vQ2 = (162 - 112) x 2
= 100
At point R,
Total energy = K.E. + P.E.
162m = ½ mvR2 + mgh
3
162m = m (½ vR2 + 10 x 9.0)
162 = ½ vR2 + 90
vR2 = (162 – 90) x 2
= 144
Example 8:
A skier of mass 60 kg slides down a smooth slope as shown in the diagram below. The height of the slope is 20 m.
(a) At the instant where the skier reaches the bottom,
(i) what is amount of energy possessed by the skier? (ii) what is the speed of the skier?
(b) State any assumptions you made in (a)(ii)
20 m Analysis:
- Slide start from rest i.e. v = 0 ms-1
(a)(i)
20 m
At the top. Total Energy = G.P.E. At the bottom, Total Energy = K.E.
Loss of G.P.E. = Gain in K.E.
Amount of Energy possessed by the skier at the end
= ∆K.E = ∆G.P.E = mgh
= (60)(10)(20)
(a)(ii)
∆K.E =
12000 =
v = = 20 ms-1
Speed of skier is 20 ms-1
2 2 1 mv 2 ) 60 ( 2 1 v 60 ) 12000 ( 2 (b)
Assume no friction and air resistance.
20 m
5 N 3 kg
5m
(a) P.E. gained = 150 J
(b) Work done by terry = 200 J
Let’s Look @ Example 4 again…
Think…..
Is there a problem here?
Principle of conservation of energy doesn’t apply?
What happened to 50 J of energy?
50 J of work done is used to overcome friction i.e. converted to heat and sound.
3. Power
Power measures the rate of doing work
or the rate at which energy is transformed.
i.e. How fast work is done.
taken
Time
change
Energy
taken
Time
Done
Work
Power
The SI unit of power is Watt (W).
One Watt simply means one Joule per second.
A crane lifts a 540kg up a vertical distance of 8m in 1 minute. Taking g to be 10 N kg-1.
Calculate
(a) the work done by the crane
(b) the power of the crane
540 kg
8 m in 1 min
(a) W = F x d
= 540(10) x 8
= 43200 J
(b) P = W / t
= 43200 / t
= 43200 / (1x60)
= 720 W
A car moves at a constant velocity of 25 ms-1 along a
straight road for 1 km. The resistive force acting on the car is 4000 N throughout the journey.
Example 10:
(a) What is the forward driving force provided by the car engine?
(b) What is the work done by the car engine?
RECALL that in “Chap 3 DYNAMICS”
Constant Velocity
Newton’s 1st Law
- Balanced Forces
- No Net Force
Newton’s 2nd Law
- a = 0 ms-2
- Fnet = ma = m(0) = 0 N
(a) Car moving at constant velocity implies that net force acting on car is zero.
Net force = Sum of all forces acting on car
0 = FE – 4000
FE = 4000 N
4000 N
Forward driving force, FE
a = 0 m/s2
Example 10:
(b) W = F x d = 4000 x 1000 = 4 x 106 J
Work done by the car engine is 4 x 106 J.
Example 10:
(c) P = W / t
= 4000000 / t = 4000000 / 40
= 1 x 105 W
Power of the car engine 1 x 105 W.
4. Efficiency
In reality, where there is friction, machines waste energy.
This energy is lost or dissipated as heat or
sound energy.
%
100
Input
Energy
Output
Energy
Useful
Efficiency
% 100 Input Power Output Power Useful EfficiencyExample 11:
A generator is able to provide 5 000J output for an input of 6 000J.
Calculate
(a) the amount of wasted energy (b) the efficiency of the generator.
(a) Wasted energy = 6000 – 5000
Example 11:
A generator is able to provide 5 000J output for an input of 6 000J.
Calculate
(b) the efficiency of the generator.
(b) Efficiency =
=
= 83.3 % (to 3 s.f.)
done over a period of time change over a
period of time
is the ability to do conversions obey Conservation of energy principle Kinetic energy
Elastic potential energy
Gravitational potential energy Chemical potential energy Heat energy Light energy Sound energy Electrical energy Magnetic energy Nuclear energy Solar energy e.g. Energy
power = = energy
time
