Question

Under the action of a force, a $2kg$ body moves such that its position $x$ as a function of time is given by $x=t^3/3$, where $x$ is in meter and $t$ in seconds. The work done by force in first two seconds is

• A. $1.6J$
• B. $16J$
• C. $160J$
• D. $1600J$

Answer 1

Answer: (B)

$16J$

Velocity of $v=\frac{dx}{dt}=t^2$

Work done by force $=K.E$ at $t=2sec$

$=\frac{1}{2}m{v}^2$

$=\frac{1}{2}m(t^2{)}^2$

$=\frac{1}{2}\times 2\times 2^4$

$=16J$ at $t=2sec$