Question

A body of mass $3.0\text{kg}$ moves under the influence of some external force such that its position s as a function of time t is given by $s=6t^3-t^2+1,$ where $s$ is in metres and $t$ is in seconds. The work done by the force in first three seconds is

• A. $18\text{J}$
• B. $1800\text{J}$
• C. $3660\text{J}$
• D. $36504\text{J}$

Answer 1

The correct option is D $36504\text{J}$

Given
$s=6t^3-t^2-1v=\frac{ds}{dt}=18t^2-2t$
At $t=0,v=0$
At $t=3s,v=18\times 9-2\times 3=156\text{m/s}$
Using work - energy theorem,
$W=\frac{1}{2}m(v_2^2-v_1^2)=\frac{1}{2}\times 3({156}^2-0)=36504\text{J}$
Hence, the correct answer is option (d).