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 d K d t =, where is a positive constant of appropriate dimensions. Which of the foll

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 $\frac{d K}{d t} = γ t$ , where $γ$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true? [2018]

Smart Achievers · Accepted Answer

(1, 2, 4)

$G \frac{d k}{d t} = γ t and k = \frac{1}{2} m V^{2}$ $∴$ $\frac{d}{d t} (\frac{1}{2} m V^{2}) = γ t$

$\Rightarrow \frac{m}{2} \times 2 V \frac{d V}{d t} = γ t$ $∴$ $m V \frac{d V}{d t} = γ t$

$∴$ $m \int_{0}^{V} V d V = γ \int_{0}^{t} t d t$ $\Rightarrow \frac{m V^{2}}{2} = \frac{γ t^{2}}{2}$

$∴$ $V = \sqrt{\frac{γ}{m}} \times t$ $i.e., V \propto t;$ $V = \frac{d s}{d t} = \sqrt{\frac{γ}{m}} t \Rightarrow s = \sqrt{\frac{γ}{m}} \cdot \frac{t^{2}}{2}$

So $V$ is proportional to $' t'$ and distance cannot be proportional to $' t'$ .

$Now F = m a = m \frac{d V}{d t} = m \frac{d}{d t} (\sqrt{\frac{γ}{m}} \times t) = m \sqrt{\frac{γ}{m}} = \sqrt{γ m} = constant$

Since force applied is constant and displacement between any two points on x-axis will also be constant, thus work done will be independent of path. Hence force is conservative in nature.

Solution

1 The force applied on the particle is constant

2 The speed of the particle is proportional to time

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

4 The force is conservative

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