A particle of mass m moves with a variable velocity v, which changes with dist

Discussion Forum : Work Power And Energy

Question -

A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line as $v = k \sqrt{x}$ ,where k is a positive constant . The work done by all the forces acting on the particle, during the first t second is

Options:

A .

\frac{m k^{4}}{t^{2}}

B .

\frac{m k^{4} t^{2}}{4}

C .

\frac{m k^{4} t^{2}}{8}

D .

\frac{m k^{4} t^{2}}{16}

Answer: Option C
:
C

GIven $v = k \sqrt{x}$ or $\frac{d x}{d t} = k \sqrt{x}$ or $x^{- \frac{1}{2}} d x = k d t$

Integrating both sides, we get

$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = k t + C$ ; Assuming x(0) = 0

Therefore, C = 0

$2 \sqrt{x} = k t \Rightarrow x = \frac{k^{2} t^{2}}{4}$ or $v = \frac{k^{2} t}{2}$

Therefore, work done,

$△ W$ = Increase in KE

= $\frac{1}{2} m v^{2} - \frac{1}{2} m (0)^{2} = \frac{1}{2} m {[\frac{k^{2} t}{2}]}^{2} = \frac{m k^{4} t^{2}}{8}$

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