Question

A constant force acts on a body of mass m at rest.The velocity V acquired during traversing a specific distance depends on m as

a ) V ' inversely proportional to ' m c ) V ' directly proportional to ' m

b ) V ' proportional to ' {- 1 / ( m ) ^{( 1 / 2 )} } d ) V ' proportional to ' ( m ) ^{( 1 / 2 )}

Answer 1

Let the specific distance traversed is 's' and the body has achieved velocity 'v' and the constant acceleration of the body is 'a'.

Then,
Work done on the body = F.s
= ma.s
now, change in KE of the body = $\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2$

=$\frac{1}{2}m{v}^2-\frac{1}{2}m{0}^2=\frac{1}{2}m{v}^2$

Now, from the law of conservation of energy the work done on the body will result in change its KE, then,

$$\begin{gathered} ma.s=\frac{1}{2}m{v}^2 \\ or, \\ a.s=\frac{1}{2}{v}^2 \\ or \\ v\propto \sqrt{a.s} \\ or \\ v\propto \sqrt{a}--------------\left(1\right) \\ and \\ \sin ce,F=ma \\ butF=cons\tan t \\ whichmeans \\ ma=cons\tan t \\ or, \\ a\propto \frac{1}{m}--------\left(2\right) \\ Fromeq\left(1\right)andeq\left(2\right)suggestthat, \\ v\propto \frac{1}{\sqrt{m}} \end{gathered}$$