Question

The displacement x of a particle of mass m moving in a straight line varies with time t as k=t^3/2 under the action of a force F where k is constant. Calculate the work done by the force

Answer 1

Answer:

Given:

Displacement varies as

$x = k {t}^{ \frac{3}{2} }$

To find:

Work done by force

Concept:

As per Work-Energy Theorem, we can say that work done is the change in Kinetic energy.

Work = ∆KE

Calculation:

$x = k {t}^{ \frac{3}{2} }$

$= > v = \frac{dx}{dt} = k \frac{d {(t)}^{ \frac{3}{2} } }{dt}$

$= > v = \frac{3}{2} k \sqrt{t}$

Now ,

$kinetic \: energy = \frac{1}{2} m {v}^{2}$

$= > ke = \frac{1}{2} m( \frac{9}{4} {k}^{2} t)$

$= > ke = \frac{9}{8} {k}^{2} t$

So final answer is :

$\boxed{ work \: done= \frac{9}{8} {k}^{2} t}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• x = kt^(3/2)

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

$\large \star {\boxed{\sf{v \: = \: \dfrac{dx}{dt}}}} \\ \\ \implies {\sf{v \: = \: \dfrac{d(kt^{\frac{3}{2}})}{dt}}} \\ \\ \implies {\sf{v \: = \: \dfrac{3}{2} k \sqrt{t}}} \\ \\ \large \star {\boxed{\sf{K.E \: = \: \dfrac{1}{2} mv^2}}} \\ \\ \implies {\sf{K.E \: = \: \dfrac{1}{2} m \bigg( \dfrac{3}{2} k \sqrt{t} \bigg)^2}} \\ \\ \implies {\sf{K.E \: = \: \dfrac{1}{2} m \bigg( \dfrac{9}{4} k^2 t \bigg)}} \\ \\ \implies {\sf{K.E \: = \: \dfrac{9}{8}k^2 t }} \\ \\ \implies {\boxed{\sf{Work \: Done \: = \: K.E \: = \: \dfrac{9}{8}k^2 t}}}$