If the point of application of a force is displaced, the force is said to do work. For a constant force F, whose point of application is given a displacement d, work done is the product of /F/ and d.

The work done by a constant force is defined as the product of the force and the distance moved by its point of application along the line of application of the force.

W= / f/d

Consider a force F displaced a distance d along is line of application AB


Power is the rate at which work is being done. For example, if work of 90j is done in 15 seconds, the power is 6J/sec. The unit of power is the Watt (W)

Example 16

On the level, a car develops a power of 60KW. if the resistance to motion is 9ooN, what is the maximum speed of the car?

Working at the same power and with the same resistance operating, what would be the maximum speed possible up an inclined plane whose slope is sin-1 , if the mass of the car is 800kg?

What is the acceleration at the time when the car is moving up the inclined plane at 40ms-1? (Take g= 10ms-2)


Let P be the power developed by the car.

Let W be the work done.

Let F be the tractive force of the car


Kinetic Energy

The work done in bringing a particle of mass m from rest to a velocity v is called the kinetic energy of the particle

If we donate Ek as the kinetic of particle of mass m reaching a velocity v from rest, then

Ek = mv2

If W is the work done in bringing the particle of mass m to a velocity v from an initial velocity u, and if s is the distance travelled in the process, then

Potential energy

Potential energy of a particle of mass m is the energy of the particle, by virtue of its position relative to a reference level. It is the work done in bringing a particle from a reference level to a height h

If we donate the potential energy by Ek then

Ep =

Since F = mg, and S = /

Ep= mgh


Law of Conversation of Energy

This is a generalization if the experience from nature and it state that energy cannot be created nor destroyed. It can only be changed from one from to another

When a particle falls from a height, it loses potential energy. This loss in potential energy is compensated for, in the gain in kinetic energy


A particle starting from rest falls freely from a height H above the ground. If g is the acceleration due to gravity, show from energy consideration that the velocity v with which the particle strikes the ground is given by the expression

v =


Let the particle have mass mkg.

Loss of potential energy = mgH.

Gain in kinetic energy =  2

from the principle of conversation of energy:

Impulse and Momentum

We recall from Newton’s secondary law of motion that for a constant force F


1) A body of mass 20kg moves a distance of 8m in the direction of line of action of force F =5N  on a smooth table find the work done

2) A particle of mass 3kg is projected vertically upward with an initial velocity of 5 m/s from the ground, calculate the potential energy at the greatest height

3) A sphere of mass 12kg and another sphere of 8kg moves toward each other with velocities 5 m/s and 3 m/s respectively find the speed of the sphere after collision

4) Calculate the loss in kinetic energy caused by the collision of the two bodies above in (3)


Reading Assignment

New Further Maths Project 2 page 245 – 257



A body at rest and of mass 8kg is acted upon by a force of 30N for 0.4 seconds, calculate the

1) impulse on the body  a) 120Ns   b) 240Ns  c) 3.2Ns  d) 12Ns

2) final speed of the body  a) 1.5m/s  b) 2.5m/s  c) 2.0m/s  d) 3m/s

3) distance covered within the time interval a) 3m  b) 30m  c) 0.3m  d) 0.354m

4) kinetic energy possessed by the body a) 6J  b) 9J  c) 12J  d) 15J

5) power of the body  a) 50W  b) 22.5W  c) 45W  d) 12.5W



1) A body of mass 6kg moves with speed 3m/s , if it is acted upon by a force of 18N for 4 seconds find the speed of the body.

2) The resistance to the motion of a cart being pushed by a man is 220N, if the man pushed the cart a distance of 10km  for 45 mins calculate (i) the work done by the man  (ii) power exerted by the man.

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