### What is 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.

### How to find Work done by Constant Force ?

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

$\displaystyle W = \vec{F}.\vec{S} $

Thus , W = F S 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

Solved Example : A particle of mass 2 kg moves under the action of a constant force $ \displaystyle \vec{F} = 5\hat{i} – 2\hat{j} $ N. If its displacement is 6i^{^} m, what is the work done by the force ?

Solution: The force acting on the body is $ \displaystyle \vec{F} = 5\hat{i} – 2\hat{j} $ N. while the displacement , $\displaystyle \vec{x} = 6\hat{i} $ m

The work done , $ W = \displaystyle \vec{F} . \vec{x}$

$ W = \displaystyle (5\hat{i} – 2\hat{j} ). (6\hat{i} ) $

W = 30 joule

Solved Example: A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration due to viscous resistance of air until half of its original height. It attains its maximum (terminal) speed and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first half and second half of the journey ?

[Density of H_{2}O = 10^{3} kg/m3]. What is the work done by the resistance force in the entire journey if its speed on reaching the ground is 10 m/s ?

Sol. r = 2 mm = 2 × 10^{-3} m, ρ = 10^{3} kg/m^{3}

Distance moved in half of the journey = s = 500/2 = 250 m.

Mass of the drop = vol. × density $ \displaystyle = \frac{4}{3}\pi r^3 \rho $

$ \displaystyle = \frac{4}{3}\times \frac{22}{7}(2\times 10^{-3})^3 \times 10^3 $

= 3.35 × 10^{-5} kg

Work done = W = F × s

= 3.35 × 10^{-5} × 9.8 × 250 = 0.082 J

Work done by the gravitational force remains same whether the drop falls under decreasing acceleration or with uniform speed. So, the work done in second half of the journey is also 0.082 J.

Energy of the drop in reaching the ground = E = mgh

= 3.35 × 10^{-5} × 9.8 × 500 = 0.164 J

(considering no resistive forces).

Kinetic energy $\displaystyle = \frac{1}{2}mv^2 = \frac{1}{2}(3.35\times 10^{-5})\times 10^2 $

= 1.675 × 10^{-3}

= 0.001675

Work done by the resistive force = 0.164 – 0.001675 = 0.1623 J

Solved Example: A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Calculate the

(i) Work done by the applied force in 10s.

(ii) Work done by friction in 10s

(iii) Work done by the net force on the body in 10s

(iv) Change in K.E. of the body in 10s and interpret your result.

Sol. m = 2kg, u = 0, F = 7N, μ= 0.1, t = 10s, W = ?

Acceleration produced by applied force =

a_{1} = F/m

= 7/2 = 3.5 m/s^{2}.

Force of friction = f = μR

= μmg = 0.1 × 2 × 9.8 = 1.96 N

Retardation produced by friction = a_{2} = (-f)/m

= (-1.96)/2

= 0.98 m/s^{2}

Net acceleration = a = a_{1} + a_{2}

= 3.50 – 0.98

= 2.52 m/s^{2}

Distance moved by the body in 10 seconds = S = ut + (1/2) at^{2}

= 0 + 1/2 × 2.52 × (10)^{2}

= 126 m.

(i) Work done by applied force = W = F S

= 7 × 126 = 882 J

(ii) Work done by the frictional force

= f × s

= 1.96 × 126

= 246.9 J

(iii) Work done by the net force

= Net force × distance

= (7 – 1.96) × 126

= 635 J

(iv) Velocity at the end of 10 s is

v = u + at

= 0 + 2.52 × 10

= 25.2 m/s

Final k.E. = (1/2) mv^{2} = 1/2 × 2 × (25.2)^{2} = 635 J

Initial K.E. = 0

Change in KE = 635 – 0

= 635 J

This shows that the change in K.E. is equal to the work done by the net force.

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 ?

### How to find Work done by a variable force ?

The equation $ W = \displaystyle \vec{F} . \vec{S}$ = FS cos θ is applicable when force remains constant but when the force is variable work is obtained by integrating $ \displaystyle \vec{F} . \vec{ds}$

Thus , $ \displaystyle W = \int \vec{F} . \vec{ds} $

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

$ \displaystyle W = \int \vec{F} . \vec{dx } = \int \vec{F}.\vec{v}dt $

Where $\vec{F}$ and $\vec{v}$ are the instantaneous force and velocity vectors.

Solved Example : A body acted upon by a force F given by , $ \displaystyle \vec{F} = -k[cos \omega t \hat{i} + sin \omega t\hat{j}] $ undergoes displacement, where the position vector of the body is given by $ \displaystyle \vec{r} = a[cos (\omega t + \alpha) \hat{i} + sin(\omega t + \alpha)\hat{j}] $ . Find the work done by the force from time t = 0 to time t = 2π /ω

Solution : The position of body is given by

$ \displaystyle \vec{r} = a[cos (\omega t + \alpha) \hat{i} + sin(\omega t + \alpha)\hat{j}] $

Its velocity is given by,

$ \displaystyle \vec{v} = \frac{\vec{dr}}{dt} = \frac{d}{dt}[a cos (\omega t + \alpha)\hat{i} + a sin(\omega t + \alpha)\hat{j}] $

$ \displaystyle = -a\omega sin (\omega t + \alpha)\hat{i} + a\omega cos(\omega t + \alpha)\hat{j} $

The power developed by this force is,

$ \displaystyle \frac{dW}{dt} = \vec{F}.\vec{v} $

= (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

$ \displaystyle = a\omega k sin\alpha \int_{0}^{\frac{2\pi}{\omega} } dt $

= ( a ω k sin α ) × 2π/ω

= 2π a k sin α

Exercise : A particle moves under a force $ \displaystyle \vec{F} = xy \hat{i} + y^2 \hat{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).

Exercise : A particle is acted upon by a force given by $ \displaystyle \vec{F}= Acos\omega t \hat{i} + B\hat{k}$ and its position vector is given by $ \displaystyle \vec{r} = a [cos\omega t \hat{i} + sin \omega t \hat{j}] + \frac{1}{2} b t^2 \hat{k}$

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