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(C) The change in momentum $\Delta p$ is equal to the area under the force-time graph.From the graph,the area from $t = 0$ to $t = 3 \ s$ is a triangl

Vedclass · Accepted Answer

$5.06$

Vedclass · Answer

(C) The change in momentum $\Delta p$ is equal to the area under the force-time graph.From the graph,the area from $t = 0$ to $t = 3 \ s$ is a triangle with base $3 \ s$ and height $4 \ N$,so Area$_1 = \frac{1}{2} 	imes 3 	imes 4 = 6 \ N \cdot s$.The area from $t = 3 \ s$ to $t = 4.5 \ s$ is a triangle with base $1.5 \ s$ and height $-2 \ N$ (since the force at $t = 4.5 \ s$ is $-2 \ N$ by similar triangles),so Area$_2 = \frac{1}{2} 	imes 1.5 	imes (-2) = -1.5 \ N \cdot s$.The total change in momentum $\Delta p = 6 - 1.5 = 4.5 \ kg \cdot m/s$.Since the block starts from rest,the final momentum $p = 4.5 \ kg \cdot m/s$.The kinetic energy $K = \frac{p^2}{2m} = \frac{(4.5)^2}{2 	imes 2} = \frac{20.25}{4} = 5.0625 \ J \approx 5.06 \ J$.

$A$ block of mass $2 \ kg$ moves freely along the $x$-axis. It is at rest at $t = 0$. The force $F(t)$ acting on the block varies with time $t$ as shown in the figure. What is the kinetic energy of the block after $4.5 \ s$ (in $J$)?

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