Answer
(a) $\mu_k = 0.078$
(b) The decrease in the bullet's kinetic energy is 248 J.
(c) The kinetic energy of the block is 0.441 J.
Work Step by Step
(a) We can use conservation of momentum to find the speed of the block just after the collision. Let $A$ be the bullet and let $B$ be the block.
$m_A~v_{A2}+m_B~v_{B2} = m_A~v_{A1}$
$v_{B2} = \frac{m_A~v_{A1}-m_A~v_{A2}}{m_B}$
$v_{B2} = \frac{(0.00400~kg)(400~m/s)-(0.00400~kg)(190~m/s)}{0.800~kg}$
$v_{B2} = 1.05~m/s$
We can find the rate of deceleration of the block.
$a = \frac{v^2-v_{B2}^2}{2x}$
$a = \frac{0-(1.05~m/s)^2}{(2)(0.720~m)}$
$a = -0.766~m/s^2$
We can use the magnitude of acceleration to find the coefficient of kinetic friction.
$F_f = ma$
$mg~\mu_k = ma$
$\mu_k = \frac{a}{g}$
$\mu_k = \frac{0.766~m/s^2}{9.80~m/s^2}$
$\mu_k = 0.078$
(b) $K_1 = \frac{1}{2}m_A~v_{A1}^2$
$K_1 = \frac{1}{2}(0.004~kg)(400~m/s)^2$
$K_1 = 320~J$
$K_2 = \frac{1}{2}m_A~v_{A2}^2$
$K_2 = \frac{1}{2}(0.004~kg)(190~m/s)^2$
$K_2 = 72.2~J$
$\Delta K = K_2-K_1$
$\Delta K = 72.2~J - 320~J$
$\Delta K = -248~J$
The decrease in the bullet's kinetic energy is 248 J.
(c) $K_B = \frac{1}{2}m_B~v_{B2}^2$
$K_B = \frac{1}{2}(0.800~kg)(1.05~m/s)^2$
$K_B = 0.441~J$
The kinetic energy of the block is 0.441 J.
