A force is applied to a particle of mass 0.1 | vedclass

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(D) The work done by the force is equal to the area under the $F-x$ graph.Area $= 	ext{Area of trapezoid} = \frac{1}{2} 	imes (	ext{sum of parallel

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$40$

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(D) The work done by the force is equal to the area under the $F-x$ graph.Area $= 	ext{Area of trapezoid} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.Area $= \frac{1}{2} 	imes (12 + (8 - 4)) 	imes 10 = \frac{1}{2} 	imes (12 + 4) 	imes 10 = \frac{1}{2} 	imes 16 	imes 10 = 80 \ J$.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,$K_i = 0$.$80 = \frac{1}{2} m v^2$.$80 = \frac{1}{2} 	imes 0.1 	imes v^2$.$160 = 0.1 	imes v^2$.$v^2 = 1600$.$v = 40 \ m/s$.

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

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