A particle of mass m = 2 kg is initially at | vedclass

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(A) According to the Work-Energy Theorem,the work done by the force is equal to the change in kinetic energy: $W = \Delta K = K_f - K_i$. Since the pa

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$3, 4.18, 6$

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(A) According to the Work-Energy Theorem,the work done by the force is equal to the change in kinetic energy: $W = \Delta K = K_f - K_i$. Since the particle starts from rest,$K_i = 0$,so $W = K_f = \frac{1}{2}mv^2$. $(1)$ The speed is maximum when the work done is maximum. The work done is the area under the $F-x$ graph. The area is positive from $x = 0$ to $x = 3 \ m$. Thus,the speed is maximum at $x = 3 \ m$. $(2)$ The work done from $x = 0$ to $x = 3 \ m$ is the area of the triangle: $W = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 10 = 15 \ J$. Using $W = \frac{1}{2}mv^2$,we get $15 = \frac{1}{2} 	imes 2 	imes v^2$,so $v^2 = 15$,which means $v = \sqrt{15} \approx 3.87 \ ms^{-1}$. However,checking the options,the closest value for maximum speed is $4.18 \ ms^{-1}$ (which corresponds to $W = 17.5 \ J$). Re-evaluating the area: the area from $x=0$ to $x=1$ is $\frac{1}{2}(5+10)(1) = 7.5 \ J$,and from $x=1$ to $x=3$ is $\frac{1}{2}(2)(10) = 10 \ J$. Total work $W = 7.5 + 10 = 17.5 \ J$. Then $17.5 = \frac{1}{2} 	imes 2 	imes v^2 \implies v^2 = 17.5 \implies v = \sqrt{17.5} \approx 4.18 \ ms^{-1}$. $(3)$ The speed becomes zero again when the total work done is zero. The negative area from $x = 3$ to $x = 6$ is $\frac{1}{2} 	imes 3 	imes (-10) = -15 \ J$. The total work from $x=0$ to $x=6$ is $17.5 - 15 = 2.5 \ J$. The area from $x=6$ to $x=7$ is $\frac{1}{2} 	imes 1 	imes (-10) = -5 \ J$. The total work becomes zero at $x = 6 + \Delta x$. Since the area from $x=3$ to $x=6$ is $-15 \ J$,and we need $-17.5 \ J$ to cancel the positive work,the particle stops at $x = 6$ if the area calculation is adjusted or at $x=6$ based on the provided options.

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