Question

A body of mass 0.5 kg is supplied with a power 'P' (in watt) which varies with time 't' (in second) as $P=3t^2+3$. If the velocity of the body at time t=0 is zero, then the velocity of the body at time t=3s is

• A. 12 ms$^{-1}$
• B. 24 ms$^{-1}$
• C. 18 ms$^{-1}$
• D. 36 ms$^{-1}$

Hint:

The Work-Energy Theorem provides a powerful link between dynamics (forces and power) and kinematics (energy and velocity). When power is a function of time, integrate the power to find the total work done, which then gives the change in kinetic energy.

Answer 1

The Correct Option is A

Step 1: Relate power, work, and kinetic energy.
The Work-Energy Theorem states that the work done ($W$) on an object is equal to the change in its kinetic energy ($\Delta KE$). \[ W = \Delta KE = KE_f - KE_i. \] Power ($P$) is the rate at which work is done, $P = \frac{dW}{dt}$. Therefore, the total work done over a time interval is the integral of power. \[ W = \int_{t_i}^{t_f} P(t) dt. \]

Step 2: Calculate the work done on the body from t=0 to t=3s.
We integrate the given power function $P(t) = 3t^2+3$. \[ W = \int_0^3 (3t^2+3) dt. \] \[ W = \left[ t^3 + 3t \right]_0^3 = (3^3 + 3(3)) - (0^3 + 3(0)) = (27+9) - 0 = 36 \text{ J}. \]

Step 3: Apply the Work-Energy Theorem to find the final velocity.
We are given that the initial velocity is zero, so the initial kinetic energy $KE_i=0$. The final kinetic energy is $KE_f = \frac{1}{2}mv_f^2$, where $m=0.5$ kg. \[ W = KE_f - KE_i. \] \[ 36 = \frac{1}{2}(0.5)v_f^2 - 0. \] \[ 36 = \frac{1}{4}v_f^2. \]

Step 4: Solve for the final velocity $v_f$.
\[ v_f^2 = 36 \times 4 = 144. \] \[ v_f = \sqrt{144} = 12 \text{ m/s}. \] \[ \boxed{v_f = 12 \text{ m/s}}. \]