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(B) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta K$.For a small displacem

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$5$

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(B) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta K$.For a small displacement $dx$,the work done is $dW = F \cdot dx$,so $F = \frac{dK}{dx}$.This means the force $F$ is equal to the slope of the kinetic energy versus position $(K-x)$ graph.We need to find the force at $x = 9 \, m$. This point lies on the line segment between $x = 6 \, m$ and $x = 10 \, m$.The coordinates of the endpoints of this segment are $(6, 20)$ and $(10, 0)$.The slope of this line is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 20}{10 - 6} = \frac{-20}{4} = -5 \, N$.The magnitude of the force is $|F| = |	ext{slope}| = |-5| = 5 \, N$.

The kinetic energy $K$ of a particle moving along the $x$-axis varies with its position $x$ as shown in the figure. The magnitude of the force acting on the particle at $x = 9 \, m$ is ............ $N$.

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