Question

A variable force F = k x2 acts on a particle which is initially at rest. Calculate thework done by the force during the displacement of the particle from x = 2 m to x = 8 m.(Assume the constant k =2 N m-2)​

Answer 1

Explanation:

The work done by the force during the displacement of the particle from x=0 m to x=4 m is 21.33 J.

Explanation:

Work done in moving a particle from one position to another is defined as the dot product of force and the displacement.

But, when the force acting on the particle is variable then, the work done is given by

W=\int \vec F\cdot d\vec rW=∫

F

⋅d

r

.

Since, the force is the function of xx only, therefore,

W=\int F\ dxW=∫F dx

Given:

F=kx^2F=kx

2where, k=1\ Nm^{-2}k=1 Nm

−2

.

Putting the values, the work done by the force during the displacement of the particle from x=0 m to x=4 m is given by

$$\begin{lgathered}W=\int F\ dx \\= \int^{x=4}_{x=0} kx^2\ dx \\= \left ( k\dfrac{x^3}{3}\right )^{x=4}_{x=0}\\ = \dfrac k3 (4^3-0^3) \\= \dfrac{64\ k}{3}\\ = \dfrac{64}{3}\ J = 21.33\ J.\end{