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 the following statements is (are) true?
A. The force applied on the particle is constant
B. The speed of the particle is proportional to time
C. The distance of the particle from the origin increases linearly with time
D. The force is conservative

Answer 1

Hint:
Here first we have to apply the kinetic energy formula and then differentiate it with respect to time. At last we apply the formula of force to get the answer.

Complete step by step solution:
Kinetic energy- The energy an object has because of its motion is kinetic energy. If we want to accelerate an object, a force must be applied.
Kinetic energy is given by-
K.E $ = \dfrac{1}
{2}m{v^2}$
 where $m$ is the mass of a body and $v$ is the velocity of a body.
Acceleration- Acceleration is a vector quantity defined as the rate at which the velocity of an object varies.
Acceleration is given by-
$a = \dfrac{v} {t}$
where $a$ is the acceleration,
$v$ is the velocity and
 $t$ is the time taken.
Velocity- Velocity is defined as the shift in the rate per unit time of displacement.
Velocity is given by-
$v = \dfrac{{\Delta x}}{t}$,
where $\Delta x$is the change in displacement and
$t$ is the time taken.
Force- A force may cause the velocity of an object with mass to shift i.e. to accelerate.
Force is given by-
$F = ma$
 where $F$ is the force ,
$m$ is the mass and
 $a$ is the acceleration.
Using these formulae and concepts we shall find the solution.
Given,
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.
We know that,
K.E $ = \dfrac{1}{2}m{v^2}$

Differentiating the above equation with respect to time.
$
  \dfrac{{dK}}
{{dt}} = mv\dfrac{{dv}}
{{dt}} = \gamma t \\
  \therefore m\int_0^v {vdv} = \gamma \int_0^t {tdt} \\
  \dfrac{{m{v^2}}}
{2} = \dfrac{{\gamma {t^2}}}
{2} \\
  v = \sqrt {\dfrac{\gamma }
{m}} t \\
   \Rightarrow v\alpha t
$

Since velocity is proportional to time, so option B is correct.
Also, we know that
$
  F = ma \\
   = m\dfrac{v}
{t} \\
   = m\sqrt {\dfrac{\gamma }
{m}} \\
   = \sqrt {\gamma m} = {\text{constant}}
$

So, option A is correct
Also, it is conservative in nature, as the force applied is constant.
So, option D is correct.
Hence, options A, B and D are correct.

Note:
We have to be careful while integrating and differentiating the equations as if we make a mistake we shall get a very different answer.