Question 315: Work, Energy and Power & Circular Motion

Question

A particle of mass $$m$$ is initially at rest at the origin. It is subjected to a force and starts moving along the $$x$$-axis. Its kinetic energy $$K$$ changes with time as $$dK/dt = \gamma t,$$ where $$\gamma $$ is a positive constant of appropriate dimensions. Which of a positive constant of appropriate dimensions. Which of the following statement is (are) true?

Accepted Answer

Given, a particle at rest is subjected to force. Motion of particle is along x-axis.
Change in kinetic energy is given as
$$K = {{dK} \over {dt}} = \gamma t$$ ..... (1)
where $$\gamma$$ is positive constant.
We know that kinetic energy is given by
$$K = {1 \over 2}m{v^2}$$...... (2)
Now, differentiate Eq. (2), we get
$${{dK} \over {dt}} = {2 \over 2}mv{{dv} \over {dt}} = mv{{dv} \over {dt}}$$
$${{dK} \over {dt}} = mv{{dv} \over {dt}}$$ ....... (3)
From Eqs. (1) and (3), we get
$$mv{{dv} \over {dt}} = \gamma t$$
$$ \Rightarrow mv\,dv = \gamma t\,dt$$ ...... (4)
Integrating Eq. (4), we get
$$\int {mv\,dv = \int {\gamma t\,dt \Rightarrow m\int {v\,dv = \gamma \int {t\,dt} } } } $$
$${{m{v^2}} \over 2} = {{\gamma {t^2}} \over 2} \Rightarrow m{v^2} = \gamma {t^2}$$
$$ \Rightarrow {v^2} = {\gamma  \over m}{t^2} \Rightarrow v = \sqrt {{\gamma  \over m}} t$$
$$ \Rightarrow v \propto t$$
Since, $$\gamma$$ and m are constant so the speed of particle is proportional to time therefore option (B) is true.
Also, $$a = {{dv} \over {dt}} \Rightarrow a = \sqrt {{\gamma  \over m}} $$ = constant
F = ma = constant
So, option (A) is true.
Also, $$v = {{dr} \over {dt}}$$
$$ \Rightarrow {{dr} \over {dt}} = \sqrt {{\gamma  \over m}} t$$
$$ \Rightarrow dr = \sqrt {{\gamma  \over m}} t\,dt$$ ...... (5)
Integrating Eq. (5), we get
$$\int {dr = \int {\sqrt {{\gamma  \over m}} t\,dt \Rightarrow r = \sqrt {{\gamma  \over m}} {{{t^2}} \over 2}} } $$
$$ \Rightarrow r \propto {t^2}$$
So, the distance of the particle from the origin increases at t
2
so option (C) is false.
A force is conservative if work done by the force in any closed loop is zero (i.e., work done is independent of the path). If force is constant (both magnitude and direction) then work done by the force in a closed loop is given by
$$W = \oint {\overrightarrow F \,.\,d\overrightarrow r  = \overrightarrow F \,.\,\oint {d\overrightarrow r  = \overrightarrow F \,.\,\overrightarrow 0 }  = 0} $$.
Note that we are able to take $$\overrightarrow F $$ out of integral sign because $$\overrightarrow F $$ is constant and $$\oint {d\overrightarrow r } $$ is zero because integration is carried out over a closed loop. The argument 'since force is constant, hence it is conservative' should be understood properly. The kinetic friction force and viscous drag force (when body moves with a terminal speed) are both constant (in magnitude) but are non-conservative. These forces cannot start the particle motion from the rest. The situation given in this problem is similar to a particle dropped from rest under the gravitational force. This force is constant, can initiate the motion given in the problem and is conservative.

Answer

The force applied on the particle is constant

Answer

The speed of the particle is proportional to time

Answer

The distance of the particle from the origin increases linearly with time

Answer

The force is conservative

Work, Energy and Power & Circular Motion

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