Question

A particle of mass m starts moving from origin along x-axis and its velocity varies with position (x) as $v=k\sqrt{x}$. The work done by force acting on it during first "t" seconds is
A. $\dfrac{m{{k}^{4}}{{t}^{2}}}{4}$
B. $\dfrac{m{{k}^{2}}t}{2}$
C. $\dfrac{m{{k}^{4}}{{t}^{2}}}{8}$
D. $\dfrac{m{{k}^{2}}{{t}^{2}}}{4}$

Answer 1

In this question we have to find the work done in terms of $t$, i.e. time. And we have given the velocity of the particle in terms of position (x) for which we have to find work done. We know that the maximum work done is simply given as the product of force and displacement. In order to find work done, we have to first find out force and displacement in terms of “t”, time.

Formula used:
$W=Fx$
Work = force $\times$ displacement

Complete step by step answer:
Given, velocity varies with position as $v=k\sqrt{x}$. We know that the velocity is also given as rate of change of position with respect to time, i.e.
$v=\dfrac{dx}{dt}$
Hence, we can write

\Rightarrow \dfrac{dx}{\sqrt{x}}=kdt $$ Integrating both sides we have, $$2\sqrt{x}=kt \\\ \Rightarrow x=\dfrac{{{k}^{2}}{{t}^{2}}}{4} $$………(1) Now, in order to find work done we have to find force, as work is given as a product of force and displacement (x). $$W=Fx$$ We know, force is given as product of mass and acceleration, i.e. $$F=ma \\\ $$ And acceleration is given as rate of change of velocity with respect to time $$F=m\dfrac{dv}{dt} \\\ $$ Substituting the value of (v) and then (x) from equation (1) $$F=m\dfrac{k}{2\sqrt{x}}\dfrac{dx}{dt} \\\ \Rightarrow F=m\dfrac{k}{2\sqrt{x}}(k\sqrt{x)} \\\ \Rightarrow F=\dfrac{m{{k}^{2}}}{2} \\\ $$ Hence, we can write $$W=Fx \\\ \Rightarrow W=\dfrac{m{{k}^{2}}}{2}x \\\ $$ Substituting value of x, we have $$W=\dfrac{m{{k}^{2}}}{2}\left( \dfrac{{{k}^{2}}{{t}^{2}}}{4} \right) \\\ \Rightarrow W=\dfrac{m{{k}^{4}}{{t}^{2}}}{8} $$ Therefore, the work done on the particle by the force acting on it for $t$ seconds is $$W=\dfrac{m{{k}^{4}}{{t}^{2}}}{8}$$ **Hence, the correct answer is option C.** **Note:** For such types of questions, one should know the basics of differentiation and integration. Here, we have calculated the maximum work done. In general, work done is given as the dot product of the force and displacement and the angle between the quantities is also considered. As it is a dot product, if force acting on the body is perpendicular to the direction of the displacement, then no work is done.