###### Chapter Content :

**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 30° E of north. What is the work done by the tug ?**

__Work done by a variable force__

__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**