Physics

Work done when force and displacement are inclined (Oblique case)
Consider a force 'F' acting at angle θ to the direction of displacement 's' as shown in fig.


Work done when force is perpendicular to Displacement
θ = 90º
W = F.S × cos 90º = F.S × 0 = 0
Thus no work is done when a force acts at right angle to the displacement.

Special Examples :
When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.
Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.
A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.

Example : 6
A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60º with the horizontal so as to move the toy cart by a distance horizontally. Calculate the work done.
Solution. Given F = 100 N, s = 3 m, = 60º.
Work done is given by
W = Fs cos θ = 100 × 2 × cos 60º
= 100 × 3 × = 150 J (cos 60º = )

Example : 7
An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.
Solution. Given W = 64,000 J; F = 8,000 N
Work done is given by W = Fs
or 64000 = 8000 × s
or s = 8 m