Question

A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is:-
(A) $\sqrt {mk} {t^{ - 1/2}}$
(B) $\sqrt {2mk} {t^{ - 1/2}}$
(C) $\dfrac{1}{2}\sqrt {mk} {t^{ - 1/2}}$
(D) $\sqrt {\dfrac{{mk}}{2}} {t^{ - 1/2}}$

Answer 1

Hint
As we know that the work done by the particle is the product of power and time, further substitute work done with change in kinetic energy and on simplifying the equation velocity of the particle will be obtained. Acceleration of the particle will be obtained from velocity equation which can be used to find the force on the particle
- $Work = Power \times Time$
- $K.E = \dfrac{1}{2}(mass){(velocity)^2}$
- $a = \dfrac{{d\left( {velocity} \right)}}{{dt}}$
- $Force = mass \times acceleration$

Complete step by step answer
Mass of the particle = m
Power (P) = k watts
Let v be the velocity of the particle at time t and u be the initial velocity of the particle
We know that work done on the particle is given by the product of power and time
$\Rightarrow Work = Power \times Time$
$\Rightarrow W = k \times t$ …..$(1)$
Also, work done is equal to change in kinetic energy of the particle
$\Rightarrow W = K.{E_{final}} - K.{E_{initial}}$
$\Rightarrow W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}$
As the particle starts from the rest, the initial velocity of the particle will be zero
$\Rightarrow W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{(0)^2}$
$\Rightarrow W = \dfrac{1}{2}m{v^2}$ …..$(2)$
From equations $(1)$ and $(2)$ , we get
$\Rightarrow \dfrac{1}{2}m{v^2} = k \times t$
$\Rightarrow {v^2} = \dfrac{{2kt}}{m}$
$\Rightarrow v = \sqrt {\dfrac{{2kt}}{m}}$
We know that the rate of change of velocity gives acceleration
$\Rightarrow a = \dfrac{{dv}}{{dt}}$
$\Rightarrow a = \dfrac{d}{{dt}}\left( {\sqrt {\dfrac{{2kt}}{m}} } \right)$
On further simplification, we get
$\Rightarrow a = \sqrt {\dfrac{{2k}}{m}} \times \dfrac{1}{2} \times {t^{ - 1/2}}$
$\Rightarrow a = \sqrt {\dfrac{k}{{2m}}} {t^{ - 1/2}}$
We know that force is defined as the product of mass and acceleration
$\Rightarrow F = m \times a$
By substituting the value of acceleration in the force formula, we get
$\Rightarrow F = m \times \sqrt {\dfrac{k}{{2m}}} {t^{ - 1/2}}$
$\Rightarrow F = \sqrt {\dfrac{{mk}}{2}} {t^{ - 1/2}}$
The force on the given particle at time t is $F = \sqrt {\dfrac{{mk}}{2}} {t^{ - 1/2}}$
The correct answer is option (D).

Note
There is an alternate way to find the velocity of the particle. Power can also be obtained by the product of force and velocity
$\Rightarrow P = F \times v = mav$
$\Rightarrow k = mv\dfrac{{dv}}{{dt}}$
$\Rightarrow vdv = \dfrac{k}{m}dt$
By integrating the above equation, we get
$\Rightarrow {v^2} = \dfrac{{2kt}}{m}$
From this equation, velocity can be found and a further process for finding acceleration and force is the same as mentioned above.