Work Done by a Variable Force - Problem L2

In our day to day life, we see many situations where we can say that work is being done by the force.

For example, here, we can say that man is doing work on the car by applying some force.

But the work done by these forces also depends upon their nature.

So it’s important to learn about the nature of these forces.

Here the forces involved are basically of two types.

$These are, 1. Constant Force 2 . Variable Force$

$Constant Force : If both magnitude, as well as direction of the force, remains constant, then we can say that force is of constant nature.$

$Variable Force: If either magnitude, direction or both changes, then we can say that force is of variable nature.$

So in this story, we will understand work done by variable force through Problem Solving.

Consider a variable force F is applied on 2kg block.

In this example, a horizontal force is used to pull the block. Here, $F = 2 t^{2} and at t = 0 s e c, v = 0$

$This force varies with time as F = 2 t^{2} Here, t is in seconds.$

Initially, at time t=0, a block is at rest.

We have to calculate the work done by this force on a block in 2 seconds.

Solution.

We know work done by a constant force is equal to the product of force and displacement.

But in this situation force is not constant.

So the question is, how to consider force in calculating work in this case.

So in this type of problem, we will consider very small displacement almost close to zero.

For infinitesimally small-displacement we can say the force is almost constant.

$So work done for this infinitely small displacement can be given as d W = F . d x$

The total work done will be the sum of infinitely small work.

$In this example we have: F = 2 t^{2} \dots \dots . . (1) But we know, F = m a \dots \dots \dots (2) Thus from (1) and (2), we get a = t^{2} \dots . (m = 2 k g)$

$Since, a = t^{2} So, \frac{d v}{d t} = t^{2} d v = t^{2} d t Integrating above relation between limit t=0 to t=t, we get, V = \frac{t ^{3}}{3}$

$Let the displacement of the block be dx from t = t to t = t +dt then, work done by the force F in this time interval dt is d w = F . d x = 2 t^{2} d x$

$Thus small work done is d w = 2 t^{2} (d x / d t) \times d t d w = 2 t^{2} (v) d t$

$Thus total work done can be calculated by integrating the small work done between limit t=0 to t=2 d w = 2 t^{2} \times \frac{t ^{3}}{3} \times d t$

Revision

If either magnitude or direction or both changes then a force is said to be variable in nature.

In case of variable force first we calculate work done by the force for infinitesimally small displacement.

Then we add all these small-small work done to get total work done throughout the total displacement.

Since this is a continuous function so we use integral operation for addition.

The End