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

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(A) 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 k

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) 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 $x=5$ (triangle below axis): $\frac{1}{2} 	imes 3 	imes (-1) = -1.5\, J$.
Area from $x=5$ to $x=8$ (triangle above axis): $\frac{1}{2} 	imes 3 	imes 3 = 4.5\, J$.
Total work done $W = 2 - 1.5 + 4.5 = 5\, J$.
Final kinetic energy $K_f = K_i + W = 25 + 5 = 30\, J$.
(Note: Re-evaluating the graph,the area from $x=5$ to $x=8$ is a triangle with base $3$ and height $3$,area $= 4.5$. The area from $x=2$ to $x=5$ is a triangle with base $3$ and height $-1$,area $= -1.5$. The area from $x=0$ to $x=2$ is a triangle with base $2$ and height $2$,area $= 2$. Total work $= 2 - 1.5 + 4.5 = 5$. Final $KE = 25 + 5 = 30$. Since $30$ is not in the options,let's re-check the graph coordinates. At $x=8$,$F=3$. At $x=5$,$F=0$. At $x=3$,$F=-1$. At $x=2$,$F=0$. At $x=0$,$F=2$. The calculation holds. Given the options,there might be a typo in the question's options,but based on the provided graph,the answer is $30\, J$. If we assume the area calculation was intended differently,$25 + 9 = 34$ (Option $A$) would imply a work of $9\, J$.)

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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