Question

The displacement of a particle of mass 1 kg on a horizontal smooth surface is a function of time given by

$x=\frac{1}{3}{t}^3$. Find out the work done by the external agent for the first one second.

• A. 1 Joules
• B. 0.5 Joules
• C. 2 Joules
• D. None of these

Answer 1

The correct option is

C

0.5 Joules

Displacement of the particle $x=\frac{1}{3}{t}^3$ $\Rightarrow dx=t^{2}dt$ ........(1)

The velocity of the particle at any instant '$t$' is $v=\frac{dx}{dt}=t^2$

The Acceleration of the particle at any instant '$t$' is, $a$ = $\frac{d^{2}x}{d{t}^2}=2t$

Therefore, work done by the force impressed is,

$w=\int Fdx\Rightarrow w=\int 1\left(2t\right)t^{2}dtfrome{q}^n\left(1\right)$

$w=\underset{0}{\overset{t=1}{\int }}\left(1\right)\left(t{)}^2\right(2t)dt=2\underset{0}{\overset{1}{\int }}{t}^{3}dt=2{\left(\frac{t^4}{4}\right)}_0^1=0.5J$