Question

A force $F$ acting on a body depends on its displacement $x$ as $F\propto x^{\frac{-1}{3}}$. The power delivered by $F$ will depend on displacement as

• A. $x^{\frac{2}{3}}$
• B. $x^{\frac{5}{3}}$
• C. $x^{\frac{1}{2}}$
• D. $x^0$

Answer 1

The correct option is D $x^0$

Given that force $F$ acting on a body depends on its displacement $x$ as $F\propto x^{\frac{-1}{3}}$

$\Rightarrow F=k{x}^{\frac{-1}{3}}...\left(i\right)$ $($as $F=ma)$

$\Rightarrow a=\frac{k}{m}{x}^{\frac{-1}{3}}$

We know that acceleration $\left(a\right)$ as a function of $\left(x\right)$ is

$a=v\frac{dv}{dx}=\frac{k}{m}{x}^{\frac{-1}{3}}$

$\Rightarrow vdv=\frac{k}{m}{x}^{\frac{-1}{3}}dx$
Integrating on both sides,
${\int }_{v_0}^{v}vdv=\frac{k}{m}{\int }_{x_0}^x{x}^{\frac{-1}{3}}dx$

$\frac{v^2-v_0^2}{2}=\frac{3k}{2m}\left(x^{\frac{2}{3}}\right)$
$\Rightarrow v^2\propto x^{\frac{2}{3}}$
$\Rightarrow v\propto x^{\frac{1}{3}}$
$\Rightarrow v=c{x}^{\frac{1}{3}}...\left(ii\right)$, here let $c$ be the constant term.

We also know that power $\left(P\right)$ = $F.v$, by $\left(i\right)$ and $\left(ii\right)$
$P=k{x}^{\frac{-1}{3}}\times c{x}^{\frac{1}{3}}$
$\Rightarrow P\propto x^0$

Hence option D is the correct answer.