A 2 kg object is initially at rest. It is ac | vedclass

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(C) The impulse $J$ is equal to the area under the force-time graph.Impulse $J = \int F \, dt = 	ext{Area under } F-t 	ext{ graph}$.Area from $t=0$

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The impulse $J$ is equal to the area under the force-time graph.Impulse $J = \int F \, dt = 	ext{Area under } F-t 	ext{ graph}$.Area from $t=0$ to $t=2 \ s$ is $10 \ N 	imes 2 \ s = 20 \ N \cdot s$.Area from $t=2$ to $t=3 \ s$ is $-10 \ N 	imes 1 \ s = -10 \ N \cdot s$.Total impulse $J_{total} = 20 - 10 = 10 \ N \cdot s$.According to the impulse-momentum theorem,$J = \Delta p = m(v_f - v_i)$.Given $m = 2 \ kg$ and $v_i = 0$,we have $10 = 2(v_f - 0)$,which gives $v_f = 5 \ m/s$.The work done by the force is equal to the change in kinetic energy:$W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.$W = \frac{1}{2} 	imes 2 	imes (5)^2 - 0 = 25 \ J$.

$A$ $2 \ kg$ object is initially at rest. It is acted upon by a force as indicated in the figure. Find the total work done by the force in $3 \ s$. (in $J$)

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