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

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(D) According to the work-energy theorem,the total work done by the force is equal to the change in kinetic energy of the particle.Work done = Area un

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) According to the work-energy theorem,the total work done by the force is equal to the change in kinetic energy of the particle.Work done = Area under the $F-x$ graph.The area under the graph is a trapezoid with parallel sides of length $12 \,m$ and $(8-4) = 4 \,m$,and height $10 \,N$.Area = $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$Area = $\frac{1}{2} 	imes (12 + 4) 	imes 10 = \frac{1}{2} 	imes 16 	imes 10 = 80 \,J$.Since the particle starts from rest,initial kinetic energy $K_i = 0$.Final kinetic energy $K_f = \frac{1}{2} m v^2 = \frac{1}{2} 	imes 0.1 	imes v^2$.By work-energy theorem,$W = K_f - K_i$.$80 = \frac{1}{2} 	imes 0.1 	imes v^2$.$80 = 0.05 	imes v^2$.$v^2 = \frac{80}{0.05} = 1600$.$v = \sqrt{1600} = 40 \,m/s$.

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

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