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(D) Given, displacement of the block, $x = 5t^2$.Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(5t^2) = 10t \,m/s$.At $t = 0 \,s$, $v_0 = 10(0) = 0 \,m/s$

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$9 / 25$

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(D) Given, displacement of the block, $x = 5t^2$.Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(5t^2) = 10t \,m/s$.At $t = 0 \,s$, $v_0 = 10(0) = 0 \,m/s$.At $t = 3 \,s$, $v_1 = 10(3) = 30 \,m/s$.At $t = 5 \,s$, $v_2 = 10(5) = 50 \,m/s$.By the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.Work done in first $3 \,s$: $W_1 = \frac{1}{2}m(v_1^2 - v_0^2) = \frac{1}{2}m(30^2 - 0^2) = \frac{1}{2}m(900)$.Work done in first $5 \,s$: $W_2 = \frac{1}{2}m(v_2^2 - v_0^2) = \frac{1}{2}m(50^2 - 0^2) = \frac{1}{2}m(2500)$.Ratio $\frac{W_1}{W_2} = \frac{900}{2500} = \frac{9}{25}$.

An object of mass $20 \,kg$ is displaced by $x = 5t^2 \,m$ (where $t$ is time) due to the application of a force. The ratio of the work done in $3 \,s$ and $5 \,s$ is:

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