Question

A particle of mass m starts moving from origin along x-axis and its velocity
varies with position (x) as The work done by force acting on it during
first "t" seconds is
​

Answer 1

Explanation:

mass = m. velocity v(x) = k √x

Acceleration a (t) = dv/dt = dv/dx * dx/dt

= k / (2√x) * v

= k² / 2

Force F(t) = m k² /2

Displacement = x(t) as the particle moves along x axis.

v(x) = dx/dt = k√x

=> dx/√x = k dt

Integrating on both sides:

=> 2 √x = k t + c

=> At t = 0, x = 0 and Hence, c= 0

=> x = k² t² / 4

Answer 2

Correct Question:-

A particle of mass $\sf{m}$ starts moving from origin along x - axis and its velocity  varies with position $\sf{x}$ as $\sf{v=kx^{\frac{1}{2}}.}$ Find the work done by force acting on it during  first $\sf{t}$ seconds.

Solution:-

The velocity varies with position as,

$\sf{\longrightarrow v=kx^{\frac{1}{2}}}$

$\sf{\longrightarrow \dfrac{dx}{dt}=kx^{\frac{1}{2}}}$

$\sf{\longrightarrow dt=\dfrac{1}{k}\,x^{-\frac{1}{2}}\ dx}$

Integrating,

$\displaystyle\sf{\longrightarrow \int\limits_0^tdt=\dfrac{1}{k}\int\limits_0^xx^{-\frac{1}{2}}\ dx}$

$\displaystyle\sf{\longrightarrow t=\dfrac{2}{k}x^{\frac{1}{2}}}$

$\displaystyle\sf{\longrightarrow x=\dfrac{1}{4}\,k^2t^2}$

Differentiating wrt $\sf{t,}$

$\displaystyle\sf{\longrightarrow v=\dfrac{1}{2}\,k^2t}$

Again differentiating wrt $\sf{t,}$

$\displaystyle\sf{\longrightarrow a=\dfrac{1}{2}\,k^2}$

Hence the work done,

$\displaystyle\sf{\longrightarrow W=\int\limits_0^x F\ dx}$

Taking $\sf{F=ma}$ and $\sf{dx=v\ dt,}$

$\displaystyle\sf{\longrightarrow W=\int\limits_0^t mav\ dt}$

$\displaystyle\sf{\longrightarrow W=\int\limits_0^t m\cdot\dfrac{1}{2}\,k^2\cdot\dfrac{1}{2}\,k^2t\ dt}$

$\displaystyle\sf{\longrightarrow W=\dfrac{1}{4}\,mk^4\int\limits_0^t t\ dt}$

$\displaystyle\sf{\longrightarrow\underline{\underline{W=\dfrac{1}{8}\,mk^4t^2}}}$