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.

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 = θ/2, W = 0

For π/2 < θ < 3π/2, work is negative

**Example 1:** A particle of mass 2 kg moves under the action of a constant force F^{->} = (5i^{^}-2j^{^} ) N. If its displacement is 6i^{^} m, what is the work done by the force ?

**Solution:** The force acting on the body is F^{->} = (5i^{^} – 2j^{^} ) N. while the displacement , x^{->} = 6i^{^} m

The work done = F^{->}.x^{->} = (5i^{^} – 2j^{^} ) . 6i^{^}

= 30 joule

**Exercise 1:** 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 ?

**Exercise 2 :** A particle moves under a force F^{->} = xyi^{^} + y^{2}j^{^} 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).