**Work**: Work is said to be done by a force when the point of application is displaced under the influence of the force.

Work is a scalar quantity and it is measured by the product of the magnitude of force and the component of displacement along the direction of force.

Work done by Constant Force :

In fact, work is the scalar product (dot product) of the force vector and the displacement vector.

Thus , W = FS cos θ

Where F and S are the magnitude of force and displacement vectors and θ is the angle between them.

For cos θ = positive, W = positive

For θ = π/2 , W = Zero

For π/2 < θ < 3π/2, work is negative

**Example :** A particle of mass 2 kg moves under the action of a constant force N. If its displacement is 6i^{^} m, what is the work done by the force ?

**Solution:** The force acting on the body is N. while the displacement , m

The work done ,

W = 30 joule

**Exercise :** A tug, exerting a pulling force of 800 N due north, tows a barge through a distance of 1 km in a direction 300 E of north. What is the work done by the tug ?

__Work done by a variable force__

The equation = FS cos θ is applicable when remains constant but when the force is variable work is obtained by

integrating

Thus ,

An example of a variable force is the spring force in which force depends on the extension x,

i.e., F = f(x)

**When the force is time dependent, we have**

where F^{->} and v^{->} are the instantaneous force and velocity vectors.

**Example :** A body acted upon by a force F given by , undergoes displacement, where the position vector of the body is given by . Find the work done by the force from time t = 0 to time t = 2π /ω

**Solution :** The position of body is given by

Its velocity is given by,

The power developed by this force is,

= (a ω k) cos (ωt) sin (ωt + α) – (aωk) sin (ωt) cos (ωt + α)

= a ω k [ sin (ωt + α) cos ωt – cos (ωt + α) sin ωt]

= a ω k sin (ωt + α – ωt)

= a ω k sin α

W = ∫ dW

= ( a ω k sin α ) × 2π/ω

= 2π a k sin α

**Exercise :** A particle moves under a force and traverses along a path y = 4x + 1. Find the work done by the force when the particle is displaced from the point P(1, 5) to Q (2, 9).

**Exercise :** A particle is acted upon by a force given by and its position vector is given by

Find the work done on the particle by the force F^{→} from time t = π/ 2ω to time t = π /ω