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(C) According to the Work-Energy Theorem, the work done by the net force is equal to the change in kinetic energy:$W = \Delta K = K_f - K_i$Given $m =

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$2$

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(C) According to the Work-Energy Theorem, the work done by the net force is equal to the change in kinetic energy:$W = \Delta K = K_f - K_i$Given $m = 1 \ kg$, $c = 1$, $v = x^\alpha$, $x_i = 0 \ m$, $x_f = 4 \ m$, and $W = 128 \ J$.At $x = 0$, $v_i = 1 	imes (0)^\alpha = 0$.At $x = 4$, $v_f = 1 	imes (4)^\alpha = 4^\alpha$.Substituting these into the Work-Energy equation:$128 = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$128 = \frac{1}{2} (1) (4^\alpha)^2 - 0$$128 = \frac{1}{2} (4^{2\alpha})$$256 = 4^{2\alpha}$Since $256 = 2^8 = (2^2)^4 = 4^4$, we have:$4^4 = 4^{2\alpha}$$2\alpha = 4$$\alpha = 2$.

$A$ ball of mass $1 \ kg$ moves in a straight line with velocity $v = c x^\alpha$, where $c = 1$ ($SI$ unit) and $\alpha$ is a constant. If the work done by the net force during its displacement from $x = 0$ to $x = 4 \ m$ is $128 \ J$, then the value of $\alpha$ is

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