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