Question

At time $t=0$ a particle starts moving along the $x-axis.$ If its kinetic energy increase uniformly with time $t,$ the net force acting on it must be proportional to ?

• A. $\sqrt{t}$
• B. $constant$
• C. $t$
• D. $\frac{1}{\sqrt{t}}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{t}}$

as K.E. is increasing uniformly with t.

hence, $K.E.=K.t$ any constant

$\frac{1}{2}m{v}^2=k.t$

$v^2=\frac{2k}{m}.t$

$v=\sqrt{\frac{2k}{m}}\cdot \sqrt{t}$ .......$\left(1\right)$

differentiate equation $\left(1\right)$ with respect to time,

$\frac{dv}{dt}=\sqrt{\frac{2k}{m}}\cdot \frac{1}{2\sqrt{t}}$

$a=\sqrt{\frac{k}{2mt}}$

$ma=\sqrt{\frac{mk}{2t}}$

$F_{net}=\sqrt{\frac{mk}{2}}\cdot \frac{1}{\sqrt{t}}$

$\therefore$ The net force acting is proportional to $\frac{1}{\sqrt{t}}$.