Question

Power supplied to a particle of mass \(2\ \text{kg}\) varies with time as \[ P=\frac{3t^2}{2}\ \text{watt}, \] where \(t\) is in seconds. If velocity at \(t=0\) is zero, the velocity at \(t=2\ \text{s}\) is

• A. \(1\ \text{m/s}\)
• B. \(2\ \text{m/s}\)
• C. \(\sqrt{2}\ \text{m/s}\)
• D. \(4\ \text{m/s}\)

Hint:

When power is given as a function of time, integrate power with respect to time to get work or change in kinetic energy.

Answer 1

The Correct Option is B

Concept: Power is the rate of doing work or the rate of change of kinetic energy: \[ P=\frac{dW}{dt}=\frac{dK}{dt} \] So total work done from \(t=0\) to \(t=2\) is: \[ W=\int_0^2 P\,dt \] This work becomes the kinetic energy of the particle.

Step 1: Given: \[ P=\frac{3t^2}{2} \] and mass: \[ m=2\ \text{kg} \] Initial velocity is zero: \[ u=0 \]

Step 2: Calculate work done from \(t=0\) to \(t=2\). \[ W=\int_0^2 \frac{3t^2}{2}\,dt \] \[ W=\frac{3}{2}\int_0^2 t^2\,dt \] \[ W=\frac{3}{2}\left[\frac{t^3}{3}\right]_0^2 \] \[ W=\frac{3}{2}\cdot \frac{8}{3} \] \[ W=4\ \text{J} \]

Step 3: This work is converted into kinetic energy. \[ W=\frac{1}{2}mv^2-\frac{1}{2}mu^2 \] Since \(u=0\), \[ W=\frac{1}{2}mv^2 \]

Step 4: Substitute values. \[ 4=\frac{1}{2}(2)v^2 \] \[ 4=v^2 \] \[ v=2\ \text{m/s} \] Therefore, \[ \boxed{2\ \text{m/s}} \]