Question

An object of mass 20 kg. is displayed by \(x = 5t^2\) m (here t is time) by the application of a force. Then the ratio of the work done in times 3 s and 5 s is

• A. \( \frac{2}{3} \)
• B. \( \frac{4}{9} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{9}{25} \)

Hint:

Work-Energy Theorem: Work done = Change in Kinetic Energy (\(\Delta KE\)).
If \(x = kt^2\), then \(v = 2kt\) and acceleration \(a=2k\) (constant).
Force \(F=ma\) is constant.
Work done \(W = Fx\). Since \(x \propto t^2\), then \(W \propto t^2\).
So, \(W_1/W_2 = (t_1/t_2)^2\).

Answer 1

The Correct Option is D

Given mass \(m=20\) kg. Displacement \(x = 5t^2\) m. Velocity \(v = \frac{dx}{dt} = \frac{d}{dt}(5t^2) = 10t \text{ m/s}\). Acceleration \(a = \frac{dv}{dt} = \frac{d}{dt}(10t) = 10 \text{ m/s}^2\). Force \(F = ma = 20 \text{ kg} \times 10 \text{ m/s}^2 = 200 \text{ N}\). (This force is constant). Work done (W) can be calculated in two ways: 1. \(W = F \cdot s\) (if F is constant and along displacement s) 2. Work-Energy Theorem: \(W = \Delta K.E. = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2\). Assuming the object starts from rest at \(t=0\), so \(v_i=0\). Then \(W = \frac{1}{2}m v_f^2\). Work done in time \(t_1 = 3\) s: Displacement at \(t_1=3\)s: \(x_1 = 5(3)^2 = 5 \times 9 = 45\) m. Velocity at \(t_1=3\)s: \(v_1 = 10(3) = 30\) m/s. Work done \(W_1 = \frac{1}{2}mv_1^2 = \frac{1}{2}(20)(30)^2 = 10 \times 900 = 9000\) J. (Alternatively, \(W_1 = F \cdot x_1 = 200 \text{ N} \times 45 \text{ m} = 9000 \text{ J}\)). Work done in time \(t_2 = 5\) s: Displacement at \(t_2=5\)s: \(x_2 = 5(5)^2 = 5 \times 25 = 125\) m. Velocity at \(t_2=5\)s: \(v_2 = 10(5) = 50\) m/s. Work done \(W_2 = \frac{1}{2}mv_2^2 = \frac{1}{2}(20)(50)^2 = 10 \times 2500 = 25000\) J. (Alternatively, \(W_2 = F \cdot x_2 = 200 \text{ N} \times 125 \text{ m} = 25000 \text{ J}\)). The question asks for the ratio of work done "in times 3s and 5s". This usually means work done *up to* 3s and work done *up to* 5s. Ratio \( \frac{W_1}{W_2} = \frac{9000 \text{ J}}{25000 \text{ J}} = \frac{9}{25} \). This matches option (d). If the question meant work done *during* the 3rd second and *during* the 5th second, the calculation would be different (Work done from t=2 to t=3, and t=4 to t=5). But "in times 3s and 5s" usually refers to the total work done from t=0 to t=3s and from t=0 to t=5s. Since \(W \propto v^2\) and \(v \propto t\), then \(W \propto t^2\). So, \(W_1/W_2 = (t_1/t_2)^2 = (3/5)^2 = 9/25\). Also, \(W \propto x\) (since F is constant) and \(x \propto t^2\), so \(W \propto t^2\). \[ \boxed{\frac{9}{25}} \]