A body of 0.5 ,kg moves along the positive X- | vedclass

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(D) According to the work-kinetic energy theorem,the work done by the force is equal to the change in kinetic energy of the body.$W = \Delta K = K_f -

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$6.8$

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(D) According to the work-kinetic energy theorem,the work done by the force is equal to the change in kinetic energy of the body.$W = \Delta K = K_f - K_i = \frac{1}{2} m (v_f^2 - v_i^2)$The work done is equal to the area under the force-displacement graph from $x=4 \,m$ to $x=8 \,m$. From the graph,at $x=4 \,m$,$F=1.5 \,N$ and at $x=8 \,m$,$F=3 \,N$.The area is a trapezoid (or a rectangle plus a triangle) with height $h = (8-4) = 4 \,m$ and parallel sides $F_1 = 1.5 \,N$ and $F_2 = 3 \,N$.$W = 	ext{Area} = \frac{1}{2} 	imes (F_1 + F_2) 	imes \Delta x = \frac{1}{2} 	imes (1.5 + 3) 	imes 4 = 2 	imes 4.5 = 9 \,J$.Given $m = 0.5 \,kg$ and $v_i = 3.16 \,ms^{-1}$.$9 = \frac{1}{2} 	imes 0.5 	imes (v_f^2 - (3.16)^2)$$9 = 0.25 	imes (v_f^2 - 9.9856)$$36 = v_f^2 - 9.9856$$v_f^2 = 45.9856$$v_f \approx 6.78 \,ms^{-1} \approx 6.8 \,ms^{-1}$.

$A$ body of $0.5 \,kg$ moves along the positive $X$-axis under the influence of a varying force $F$ (in newton) as shown below. If the speed of the object at $x=4 \,m$ is $3.16 \,ms^{-1}$,then its speed at $x=8 \,m$ is ................. $\,ms^{-1}$.

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