A force acts on a body due to which its position varies as . Work done by this force in first two seconds is

JEE MainPhysicsModerateWork Power & Energy

A force
$\overrightarrow{F} = \left( 3t\widehat{i} + 5\widehat{j} \right)N$ acts
on a body due to which its position varies as
$\overrightarrow{s} = \left( 2t^{2}\widehat{i} - 5\widehat{j} \right)$ .
Work done by this force in first two seconds is

Show answer and solution

W = \int\overline{F} \cdot d\overline{s} = \int(3t\ \widehat{i} + 5\widehat{j}).(4t\ dt\ \widehat{i})

= \int_{0}^{2}\mspace{2mu} 12t^{2}dt = \frac{12\left\lbrack t^{3} \right\rbrack_{0}^{2}}{3} = 32\ J

Source: Grass-G / P-06 Physics paper 14 October.docx

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