The v-t graph of an object of mass 1,text{kg} | vedclass

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(D) According to the work-energy theorem,the net work done on an object is equal to the change in its kinetic energy: $W_{	ext{net}} = \Delta K = \fr

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) According to the work-energy theorem,the net work done on an object is equal to the change in its kinetic energy: $W_{	ext{net}} = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$.From the graph,the initial velocity at $t = 0\,	ext{s}$ is $v_i = 0\,	ext{m/s}$,and the final velocity at $t = 30\,	ext{s}$ is $v_f = 0\,	ext{m/s}$.Therefore,$W_{	ext{net}} = \frac{1}{2} 	imes 1 	imes (0^2 - 0^2) = 0\,	ext{J}$. Thus,statement $(A)$ is correct.Average acceleration is defined as $a_{	ext{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$.Here,$v_f = 0\,	ext{m/s}$ and $v_i = 0\,	ext{m/s}$ over the interval $t = 0$ to $t = 30\,	ext{s}$.So,$a_{	ext{avg}} = \frac{0 - 0}{30 - 0} = 0\,	ext{m/s}^2$. Thus,statement $(B)$ is correct.Since $F_{	ext{avg}} = m 	imes a_{	ext{avg}}$ and $a_{	ext{avg}} = 0$,the average force $F_{	ext{avg}} = 1 	imes 0 = 0\,	ext{N}$. Thus,statement $(C)$ is correct.Since all statements are correct,the correct option is $(D)$.

The $v-t$ graph of an object of mass $1\,\text{kg}$ is shown. Which of the following statements is correct?

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