A 10 kg brick moves along an $x$-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from $x = 0$ to $x = 8.0$m? [image:6_image25.emf]

Question

Dr. Abubacker · Accepted Answer

(b) According to the graph the acceleration $a$ varies linearly with the coordinate $x$. We may write $a = \alpha x$, where $\alpha$ is the slope of the graph.
From the graph
$\alpha = \frac{20}{8}mg_{0} = 2.5\ s^{- 2}$
The force on the brick is in the positive $x$-direction and according to Newton’s second law, its magnitude is given by
$F = \frac{a}{m} = \frac{\alpha}{m}x$
If $x_{f}$ is the final coordinate, the work done by the force is
$W = \int_{0}^{x_{f}}{F\ dx = \frac{a}{m}\int_{0}^{x_{f}}{x\ dx}}$
$= \frac{\alpha}{2m}x_{f}^{2} = \frac{2.5}{2 	imes 10} 	imes (8)^{2}$
$= 8\ J$

Q. A 10 kg brick moves along an $x$-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from $x = 0$ to $x = 8.0$m?

Explanation

Reviewed by Dr. Abubacker

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