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(B) According to the work-energy theorem,the work done by the net force on an object is equal to the change in its kinetic energy.Let the initial velo

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$2$

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(B) According to the work-energy theorem,the work done by the net force on an object is equal to the change in its kinetic energy.Let the initial velocity be $v_0$. The initial kinetic energy is $K_i = \frac{1}{2}mv_0^2$.When the velocity becomes half $(v = v_0/2)$,the kinetic energy is $K_f = \frac{1}{2}m(v_0/2)^2 = \frac{1}{8}mv_0^2$.The work done by the resistive force $F$ over a distance $d = 6 \ cm$ is:$W = K_f - K_i = \frac{1}{8}mv_0^2 - \frac{1}{2}mv_0^2 = -\frac{3}{8}mv_0^2$.Since $W = F \cdot d$,we have $F \cdot 6 = -\frac{3}{8}mv_0^2$,which gives $F = -\frac{1}{16}mv_0^2$.Now,let the bullet travel an additional distance $d_1$ before coming to rest. The initial kinetic energy for this stage is $\frac{1}{8}mv_0^2$ and the final kinetic energy is $0$.Applying the work-energy theorem again:$F \cdot d_1 = 0 - \frac{1}{8}mv_0^2$.Substituting $F = -\frac{1}{16}mv_0^2$:$(-\frac{1}{16}mv_0^2) \cdot d_1 = -\frac{1}{8}mv_0^2$.Solving for $d_1$,we get $d_1 = \frac{1/8}{1/16} = 2 \ cm$.

$A$ bullet fired from a gun into a large wooden block loses half of its velocity after traveling $6 \ cm$ inside the block. How much further distance (in $cm$) will it travel before coming to rest?

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