Question

A force f=kx2 acts on a particle at an angle of 60 degrees with the x axis, the work done in displacing the particle from x1 to x2 is?

Answer 1

force acting on the particle is given as

F = k x²

small work done in displacing the particle through distance "dx" is given as

dW = F dx Cos60

dW = (0.5)  k x² dx

total work done in displacing the particle from "x₁" to "x₂" is given as

W = $\int_{x_{1}}^{x_{2}}$ dW

W = $\int_{x_{1}}^{x_{2}}$ (0.5)  k x² dx

W = (0.5) (k/3) (x³₂ - x³₁)

Answer 2

We know the formula of work done.

dW=F.dx

dW=Fdxcos(theta), Put the given values as below:

dW=kx^2cos(60 degree)dx

dW=1/2*kx^2dx

Integrating the above equation from x1 to x2, we get

W=(1/6kx^3) from x1 to x2

Therefore, the work done would be 1/6k[x2^3-x1^3]