The force vec F = Fhat i on a particle of mas | vedclass

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(D) Initial kinetic energy $K_i = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 2 	imes (5)^2 = 25\, J$.According to the work-energy theorem, the change in ki

Vedclass · Accepted Answer

$29.5$

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(D) Initial kinetic energy $K_i = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 2 	imes (5)^2 = 25\, J$.According to the work-energy theorem, the change in kinetic energy is equal to the work done by the force, which is the area under the $F-x$ graph.Work done $W = \int_{0}^{8} F dx = 	ext{Area under the curve}$.Area from $x=0$ to $x=2$ (triangle above axis): $\frac{1}{2} 	imes 2 	imes 2 = 2\, J$.Area from $x=2$ to $x=5$ (trapezium below axis): $\frac{1}{2} 	imes (3 + 1) 	imes (-1) = -2\, J$.Area from $x=5$ to $x=8$ (triangle above axis): $\frac{1}{2} 	imes 3 	imes 3 = 4.5\, J$.Total work done $W = 2 - 2 + 4.5 = 4.5\, J$.Final kinetic energy $K_f = K_i + W = 25 + 4.5 = 29.5\, J$.

The force $\vec F = F\hat i$ on a particle of mass $2\, kg$,moving along the $x$-axis is given in the figure as a function of its position $x$. The particle is moving with a velocity of $5\, m/s$ along the $x$-axis at $x = 0$. What is the kinetic energy of the particle at $x = 8\, m$?

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