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A body of mass 3 kg is under a constant force which causes a displacement $s$ in metres in it, given by the relation $s = \dfrac{1}{3}t^2$, where $t$ is in seconds. Work done by the force in 2 seconds is :-
A
$\dfrac{5}{19}$ J
B
$\dfrac{3}{8}$ J
C
$\dfrac{8}{3}$ J
D
$\dfrac{19}{5}$ J
Detailed Solution
By the work-energy theorem: $W = \Delta KE = \dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2$
Now $s = \dfrac{1}{3}t^2 \Rightarrow v = \dfrac{ds}{dt} = \dfrac{2}{3}t$
At $t = 0$: $v_1 = 0$
At $t = 2$ s: $v_2 = \dfrac{2}{3} \times 2 = \dfrac{4}{3}$ m s$^{-1}$
Therefore $W = \dfrac{1}{2} \times 3 \times \left(\dfrac{4}{3}\right)^2 - \dfrac{1}{2} \times 3 \times (0)^2$
$W = \dfrac{3}{2} \times \dfrac{16}{9} = \dfrac{8}{3}$ J
Check: $a = \dfrac{2}{3}$ m s$^{-2}$, $F = ma = 2$ N, $s = \dfrac{4}{3}$ m, so $W = Fs = \dfrac{8}{3}$ J.
Now $s = \dfrac{1}{3}t^2 \Rightarrow v = \dfrac{ds}{dt} = \dfrac{2}{3}t$
At $t = 0$: $v_1 = 0$
At $t = 2$ s: $v_2 = \dfrac{2}{3} \times 2 = \dfrac{4}{3}$ m s$^{-1}$
Therefore $W = \dfrac{1}{2} \times 3 \times \left(\dfrac{4}{3}\right)^2 - \dfrac{1}{2} \times 3 \times (0)^2$
$W = \dfrac{3}{2} \times \dfrac{16}{9} = \dfrac{8}{3}$ J
Check: $a = \dfrac{2}{3}$ m s$^{-2}$, $F = ma = 2$ N, $s = \dfrac{4}{3}$ m, so $W = Fs = \dfrac{8}{3}$ J.