The kinetic energy (E_k) of a particle moving | vedclass

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(D) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta E_k$.For an infinitesima

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$-5 \hat{i} \ N$

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(D) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta E_k$.For an infinitesimal displacement $dx$,the work done is $dW = F \cdot dx$.Therefore,$F = \frac{dE_k}{dx}$.This means the force acting on the particle is equal to the slope of the $E_k$ versus $X$ graph.We need to find the force at $X = 10 \ m$. This point lies on the line segment between $X = 8 \ m$ and $X = 12 \ m$.The coordinates of the endpoints of this segment are $(8, 40)$ and $(12, 20)$.The slope of this line is $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20 - 40}{12 - 8} = \frac{-20}{4} = -5 \ N$.Since the slope is constant between $X = 8 \ m$ and $X = 12 \ m$,the force at $X = 10 \ m$ is $-5 \hat{i} \ N$.

The kinetic energy $(E_k)$ of a particle moving along the $X$-axis varies with its position $(X)$ as shown in the figure. The force acting on the particle at $X = 10 \ m$ is

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