A particle of mass 0.1 , kg is subjected to a | vedclass

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(D) The work done by the force is equal to the area under the force-displacement $(F-x)$ graph.Area = Area of the trapezium with parallel sides $4 \,

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$40$

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(D) The work done by the force is equal to the area under the force-displacement $(F-x)$ graph.Area = Area of the trapezium with parallel sides $4 \, m$ and $12 \, m$ and height $10 \, N$.Area = $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$Area = $\frac{1}{2} 	imes (12 + 4) 	imes 10 = \frac{1}{2} 	imes 16 	imes 10 = 80 \, J$.According to the work-energy theorem,the work done is equal to the change in kinetic energy.$W = \Delta K = K_f - K_i$Since the particle starts from rest,$K_i = 0$.$80 = \frac{1}{2} m v^2$$80 = \frac{1}{2} 	imes 0.1 	imes v^2$$v^2 = \frac{80 	imes 2}{0.1} = 1600$$v = \sqrt{1600} = 40 \, m/s$.

$A$ particle of mass $0.1 \, kg$ is subjected to a force which varies with distance as shown in the figure. If it starts its journey from rest at $x = 0$,its velocity at $x = 12 \, m$ is .......... $m/s$.

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