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(B) The resistive force is given by $F = -k v^{\frac{1}{3}}$.Using Newton's second law, $m a = -k v^{\frac{1}{3}}$, so the retardation is $a = -\frac{

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$\frac{5}{3}$

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(B) The resistive force is given by $F = -k v^{\frac{1}{3}}$.Using Newton's second law, $m a = -k v^{\frac{1}{3}}$, so the retardation is $a = -\frac{k}{m} v^{\frac{1}{3}}$.Since $a = v \frac{dv}{dx}$, we have $v \frac{dv}{dx} = -\frac{k}{m} v^{\frac{1}{3}}$.Rearranging the terms, we get $v^{1 - \frac{1}{3}} dv = -\frac{k}{m} dx$, which simplifies to $v^{\frac{2}{3}} dv = -\frac{k}{m} dx$.Integrating both sides from initial velocity $v_0$ to final velocity $0$ over a distance $s$:$\int_{v_0}^{0} v^{\frac{2}{3}} dv = -\frac{k}{m} \int_{0}^{s} dx$.Evaluating the integral: $\left[ \frac{v^{\frac{5}{3}}}{\frac{5}{3}} \right]_{v_0}^{0} = -\frac{k}{m} s$.This gives $-\frac{3}{5} v_0^{\frac{5}{3}} = -\frac{k}{m} s$.Thus, $s = \frac{3m}{5k} v_0^{\frac{5}{3}}$, which implies $s \propto v_0^{\frac{5}{3}}$.Comparing this with $s \propto v_0^\beta$, we find $\beta = \frac{5}{3}$.

$A$ bullet enters a piece of wood with velocity $v_0$ and the resistive force acting on the bullet in the wood is proportional to $v^{\frac{1}{3}}$. If the total distance travelled by the bullet is proportional to $(v_0)^\beta$, then the value of $\beta$ is

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