Question

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

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

Answer 1

Answer: (A)

$\frac{1}{2} \frac{mv^2}{t_1^2} t^2$

Answer 2

Velocity of particle accelerating uniformly in time $t_1$
$v=at_1 \Rightarrow a =\frac{v}{t_1}$
Velocity of particle in time $t$
$v'=at =\frac{vt}{t_1}$
According to work-energy theorem,
work done = change in kinetic energy
i,e $W_{ext}=\Delta K$
or $W_{ext}=\frac{1}{2}mv'^2-0$
$=\frac{1}{2}m \left(\frac{vt}{t_1}\right)^2$
$=\frac{1}{2}\frac{mv^2}{t^2_1}t^2$