Question

An object of mass \(20\ \text{kg}\) is displaced by \[ x=5t^2\ \text{m} \] (where \(t\) is time) by the application of a force. Then the ratio of the work done in times \(3\ \text{s}\) and \(5\ \text{s}\) is

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

Hint:

If displacement is given as a function of time, first find velocity using \[ v=\frac{dx}{dt}. \] Then use \[ W=\Delta K=\frac{1}{2}mv^2 \] to determine the work done.

Answer 1

The Correct Option is D

Step 1: Find acceleration and work done.
\( x = 5t^2 \Rightarrow v = 10t,\; a = 10\text{ m/s}^2 \). Work done \( = \frac{1}{2}mv^2 = \frac{1}{2}(20)(10t)^2 = 1000t^2 \).

Step 2: Take ratio at t=3 s and t=5 s.
\( \frac{W_3}{W_5} = \frac{1000(3)^2}{1000(5)^2} = \frac{9}{25} \)

\[ \boxed{\frac{W_3}{W_5} = \frac{9}{25}} \]