Question

A force $F = K{x^2}$ acts on a particle at an angle of $60^\circ$ with the x-axis, the work done in displacing the particle from ${x_1}$ to ${x_2}$ will be-
A. $\dfrac{{K{x^2}}}{2}$
B. $\dfrac{K}{2}\left( {x_2^2 - x_1^2} \right)$
C. $\dfrac{K}{6}\left( {x_2^3 - x_1^3} \right)$
D. $\dfrac{K}{3}\left( {x_2^3 - x_1^3} \right)$

Answer 1

Use the formula for the small work done for the small displacement of the particle. This formula gives the relation between the force acting on the particle, displacement of the particle and the angle between the directions of force and displacement of the particle. Substitute the given values of force and angle between the force and displacement of the particle and integrate this equation between the given displacement limits to determine the total work done.
Formula used:
The formula for small work done $dW$ by the particle is given by
$dW = Fdx\cos \theta$ …… (1)
Here, $F$ is the force acting on the particle, $dx$ is the small displacement of the particle and $\theta$ is the angle between the direction of force and direction of displacement.

Complete step by step answer:
We have given that the force acting on the particle is
$F = K{x^2}$
The force between the directions of force and displacement is $60^\circ$.
$\theta = 60^\circ$
We have asked to determine the work done in moving the particle from ${x_1}$ to ${x_2}$.
We can determine the required work done using equation (1).
Substitute $K{x^2}$ for $F$ and $60^\circ$ for $\theta$ in equation (1).
$dW = K{x^2}dx\cos 60^\circ$
$\Rightarrow dW = K{x^2}dx\left( {\dfrac{1}{2}} \right)$
$\Rightarrow dW = \dfrac{1}{2}K{x^2}dx$
Now, we can determine the total work done for the displacement of the particle from ${x_1}$ to ${x_2}$.
Let us integrate both sides of the above equation to determine the total work done for the
displacement of the particle from ${x_1}$ to ${x_2}$.
$\Rightarrow \int {dW} = \int_{{x_1}}^{{x_2}} {\dfrac{1}{2}K{x^2}dx}$
$\Rightarrow W = \dfrac{1}{2}K\int_{{x_1}}^{{x_2}} {{x^2}dx}$
$\Rightarrow W = \dfrac{1}{2}K\left[ {\dfrac{{{x^3}}}{3}} \right]_{{x_1}}^{{x_2}}$
$\Rightarrow W = \dfrac{1}{2}K\left[ {\dfrac{{x_2^3}}{3} - \dfrac{{x_1^3}}{3}} \right]$
$\Rightarrow W = \dfrac{K}{6}\left( {x_2^3 - x_1^3} \right)$
Therefore, the work done in the displacement of the particle from ${x_1}$ to ${x_2}$ is
$\dfrac{K}{6}\left( {x_2^3 - x_1^3} \right)$.
Hence, the correct option is C.

Note: The students should not forget to take the cosine of the angle between the force and displacement of the particle as the work done is the dot product of the force on the particle and displacement of the particle. If one forgets to use cosine of the angle between the force and displacement, the final answer will be incorrect. So, one should be careful while using the formula for the work done.