LECTURE SP 015 LESSON 5

CHAPTER 5

WORK, ENERGY AND POWER
5.2 a,

5.2 c (Introduce theorem) and
5.3 a

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CHAPTER 5 : WORK, ENERGY AND POWER L5
(1 HOUR)

Learning Outcome:

5.2 Energy and Conservation of Energy

a) State the principle of Conservation of Energy.
b) Apply principle of Conservation of Energy

(mechanical energy and heat energy due to
friction).
c) State and apply the work-energy theorem.

5.3 Power

a) Define and use average power and instantaneous
power.

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5.2 Energy and Conservation of energy
Energy (Mechanical Energy)
o is defined as the system’s ability to do work.
o The S.I. unit for energy is same to the unit of

work (joule, J).
o is a scalar quantity.

3

1. Potential Energy, U

o is defined as the energy stored in a body or
system because of its position, shape and state.

1.1 Gravitational potential energy, Ug
o is defined as the energy stored in a body or system

because of its position.

o Equation : U : gravitational potential energy

U g  mgh m : mass of a body
U  mgh g : acceleration due to gravity

where h: height of a body from the initial position

o The gravitational potential energy depends only on

the height of the object above the surface of the 4
Earth.

1.2 Elastic potential energy, Us

o is defined as the energy stored in in elastic materials

as the result of their stretching or compressing.

o Springs are a special instance of device which can
store elastic potential energy due to its compression
or stretching.

o Hooke’s Law states “the restoring force, Fs of spring is
directly proportional to the amount of stretch or

compression (extension or elongation), x if the limit of

proportionality is not exceeded”

o where Fs  x Fs  kx

Fs : the restoring force of spring
k : the spring constant or force constant
x : the amount of stretch or compression (x f -xi ) 5

o Negative sign in the equation indicates that the
direction of Fs is always opposite to the direction of
the amount of stretch or compression (extension), x.

 Case 1: (Vertical spring)

The spring is hung vertically and its is stretched by a
suspended object with mass, m as shown in figure below.

Initial position  The spring is in equilibrium
Final position Fs
x thus
https://youtu.be/0BObd3DsNFM
Fs  W  mg

   6
W mg

 Case 2: (Horizontal spring)

The spring is attached to an object and it is stretched

and compressed by a force, F as shown in below.

Fs is negative Fs 
x is positive F

x The spring is in

equilibrium, hence

x0 Fs  F

Fs  0 F = applied force
x0

(Equilibrium position)

 x0
F
 Fs is positive

x Fs x is negative

7

 Caution: For calculation, use :

Fs  F  kx where F : applied force

◦ The unit of k is kg s2 or N m1

 From the Hooke’s law (without “ ve” sign), a restoring

force, Fs against extension of the spring, x graph is
shown in the graph below.

Fs W  area under the Fs  x graph
F

W  1 Fx1 W  1 kx1 x1
2 2

0 x1 x W  1 kx12 Us
2
https://youtu.be/UyO_wNwc4s0
8

o From the F-x graph the work done to stretch/
compress the spring is equal to the elastic potential
energy US stored in the spring.

o Therefore the equation of elastic potential energy,
Us for compressing or stretching a spring is given by

W Us  1 kx2  1 Fx
2 2

9

2. Kinetic Energy, K

o is defined as the energy of a body due to its motion.

 Equation :

K  1 mv2
2

K : kinetic energy of a body

where m : mass of a body

v :speed of a body

10

Principle of Conservation of Energy

o states “in an isolated (closed) system, the total energy
of that system is constant”.

o According to the principle of conservation of energy,

we get  Ei   E f

The initial of total energy = the final of total energy

E U K

https://youtu.be/SrqPNXLNCoI

K  U  K  U https://physics.info/energy-
i i f f conservation/spiderman.mp4

1 mu2  mghi  1 kxi2  1 mv2  mghf  1 kx2f 11
2 2 2 2

Example 1

Find the amount of compression of the spring if the ball
does free fall from 1 m and compresses the spring.
(0.25 m)

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Work-Energy Theorem

“The work done by the nett force on a body equals
the change in the body’s kinetic energy”

o Consider a block with mass, m moving along the horizontal
surface (frictionless) under the action of a constant nett force,

Fnett undergoes a displacement, s.

m Fnett


s

 F  Fnett  ma (1)

By using an equation of linear motion: v2  u 2  2as

a  v2 u2 (2) 13
2s

Substitute equation (2) into (1),

m v 2 u 2 
2s
Fnett 

Fnett s  1 mv2  1 mu2  Kf  Ki
2 2

Therefore Wnett  K

Fnett s cos  1 mv2  1 mu 2
2 2

14

5.3 Power

Average power, Pav

o When a quantity of work W is done during a time
interval t, the average power Pav is defined as

Pav  W
t

o Average power is defined as the work done or
energy per unit time.

o The average power is often simply called "power"

Pave  PW  Fs cos  K  mgh
t t t
t

15

Instantaneous Power , P

The instantaneous power, P is defined as the
instantaneous rate of doing work, which can be
write as

P  limit W  dW
t0 t dt

 Consider an object that is moving at a constant velocity v
along a frictionless horizontal surface and is acted by a

constant force, F directed at angle  above the horizontal as

shown below. The object undergoes a displacement of ds.



F 
ds

16

Therefore the instantaneous power, P is given by

P  dW and dW  F cosθ ds

dt

P  F cosθds and v  ds
dt
dt
OR P    
P  Fv cosθ F v

where v : magnitude of velocity

F : magnitude of force  and 
θ : the angle between F v

o Power is a scalar quantity.

o The S.I. unit of the power is kg m2 s3 or J s1 or Watt (W).

o Unit conversion : Watt (W) to horsepower (hp). 17

1 hp  746 W

Example 2
A 1990 kg car accelerates uniformly from
rest to a speed of 17.1 m/s in 4.21 s.
Calculate
a) the average power and (6.91 x 103 W)
b) the instantaneous power delivered by the

net force to the car at 2 s. (6.56 x 104 W)

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