JEE Challenger
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PhyMedAdv 2018 • Paper 2

Analysis of Motion and Kinetic Energy Rate for a Particle

Consider a particle having mass mm that is at rest at the origin initially. It experiences a force and begins moving along the xx-axis. Its kinetic energy KK varies with time according to dKdt=γt\frac{dK}{dt} = \gamma t, where γ\gamma is a positive constant possessing appropriate dimensions. Which among the following statements is (are) correct?

Options

A

The force applied to the particle is constant

Correct
B

The speed of the particle varies directly with time

Correct
C

The distance of the particle from the origin increases linearly as time passes

D

The applied force is conservative in nature

Correct

Topics & Concepts

Work, Energy and PowerWork-Energy TheoremNewton's LawsKinematics

Step-by-Step Solution

To determine which of the given statements are correct, we analyze the motion of the particle step-by-step.

  1. Relation for Kinetic Energy (KK) and Speed (vv): The rate of change of kinetic energy with respect to time is given as: dKdt=γt\frac{dK}{dt} = \gamma t

    Integrating both sides with respect to time tt from the initial state at rest (t=0,K=0t = 0, K = 0): ∫0KdK′=∫0tγt′ dt′\int_0^K dK' = \int_0^t \gamma t' \, dt' K=12γt2K = \frac{1}{2} \gamma t^2

    Since kinetic energy is defined as K=12mv2K = \frac{1}{2} m v^2: 12mv2=12γt2  ⟹  v=γm t\frac{1}{2} m v^2 = \frac{1}{2} \gamma t^2 \implies v = \sqrt{\frac{\gamma}{m}} \, t

    Therefore, the speed of the particle vv is directly proportional to time tt: v∝tv \propto t This proves that Option (B) is correct.

  2. Force Applied on the Particle (FF): The acceleration aa of the particle is the time derivative of its velocity: a=dvdt=γma = \frac{dv}{dt} = \sqrt{\frac{\gamma}{m}}

    Using Newton's second law, the magnitude of the force applied to the particle is: F=ma=mγm=mγF = m a = m \sqrt{\frac{\gamma}{m}} = \sqrt{m \gamma}

    Since mm and γ\gamma are positive constants, the force FF is constant in time and position. This proves that Option (A) is correct.

  3. Distance from the Origin (xx): Since the particle moves along the xx-axis from the origin (x=0x = 0 at t=0t = 0): v=dxdt=γm tv = \frac{dx}{dt} = \sqrt{\frac{\gamma}{m}} \, t

    Integrating with respect to time gives the position: x=∫0tγm t′ dt′=12γm t2x = \int_0^t \sqrt{\frac{\gamma}{m}} \, t' \, dt' = \frac{1}{2} \sqrt{\frac{\gamma}{m}} \, t^2

    The distance increases quadratically with time (x∝t2x \propto t^2), not linearly. Therefore, Option (C) is incorrect.

  4. Nature of the Applied Force: A constant force acting on a particle moving in one dimension can be derived from a potential energy function U(x)U(x): F=−dUdx  ⟹  U(x)=−Fx+CF = -\frac{dU}{dx} \implies U(x) = -Fx + C

    Since the work done by a constant force in one-dimensional motion is path-independent, the force is conservative. This proves that Option (D) is correct.

Correct Options: (A), (B), and (D)

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