Concept of Work

Chapter Content :
1. Work done by a Constant Force & Variable Force
2. Conservative & Non-conservative Forces
3. Work done by Spring Force
4. Kinetic Energy , Work Energy Theorem
5. Conservation of Energy & Significance of potential energy
6. Instantaneous Power & average power

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 \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

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

The equation W = \displaystyle \vec{F} . \vec{S} = FS cos θ is applicable when 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 F-> and v-> are the instantaneous force and velocity vectors.

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 F from time t = π/ 2ω to time t = π /ω

Next Page →