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 dKdt=γt, where γ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

d K d t = y t as K = 1 2 m v 2 ∴ d K d t = m v d v d t = y t ∴ m ∫ 0 v v d v = y ∫ 0 t t d t ⇒ m v 2 2 = y t 2 2 ⇒ v = y m t ⇒ a = d v d t = y m = Constant and S = 1 2 a t 2 = 1 2 y m t 2 Since F = m a ∴ F = m y m = y m = Constant Since force applied is constant, it is conservative in nature.

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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?

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The force applied on the particle is constant

The speed of the particle is proportional to time

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

The force is conservative

answer is A, B, D.

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Detailed Solution

$\begin{array}{l} \frac{d K}{d t} = y t as K = \frac{1}{2} m v^{2} \\ ∴ \frac{d K}{d t} = m v \frac{d v}{d t} = y t \\ ∴ m \int_{0}^{v} v d v = y \int_{0}^{t} t d t \\ \Rightarrow \frac{m v^{2}}{2} = \frac{y t^{2}}{2} \\ \Rightarrow v = \sqrt{\frac{y}{m}} t \\ \Rightarrow a = \frac{d v}{d t} = \sqrt{\frac{y}{m}} = Constant \\ and S = \frac{1}{2} a t^{2} = \frac{1}{2} \sqrt{\frac{y}{m}} t^{2} \\ Since F = m a \\ ∴ F = m \sqrt{\frac{y}{m}} = \sqrt{y m} = Constant \end{array}$

Since force applied is constant, it is conservative in nature.

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