Question

A particle accelerating uniformly has velocity $v$ at time $t_1$. What is work done in time $t$?

• A. $\frac{1}{2}\frac{m{v}^2}{t_1^2}{t}^2$
• B. $\frac{1}{2}{\left(\frac{mv}{t_1}\right)}^2{t}^2$
• C. $\frac{m{v}^2}{t_1^2}{t}^2$
• D. $\frac{2m{v}^2}{t_1^2}{t}^2$

Answer 1

Answer: (A)

Velocity of particle accelerating uniformly in time $t_1$
$v=a{t}_1\Rightarrow a=\frac{v}{t_1}$
Velocity of particle in time $t$
$v^{\prime }=at=\frac{vt}{t_1}$
According to work-energy theorem,
work done = change in kinetic energy
i,e $W_{ext}=\Delta K$
or $W_{ext}=\frac{1}{2}m{v}^{\prime 2}-0$
$=\frac{1}{2}m{\left(\frac{vt}{t_1}\right)}^2$
$=\frac{1}{2}\frac{m{v}^2}{t_1^2}{t}^2$