Answer
(a) Angular momentum is conserved.
(b) $\omega = 11.4~rad/s$
(c) The kinetic energy of the block increased by 0.0274 J.
(d) Work = 0.0274 J
Work Step by Step
(a) The force is directed along the cord so the torque is zero. Therefore, angular momentum is conserved.
(b) We can use conservation of angular momentum to solve this question.
$L_2 = L_1$
$m~\omega_2~r_2^2 = m~\omega_1~r_1^2$
$\omega_2 = \frac{\omega_1~r_1^2}{r_2^2}$
$\omega_2 = \frac{(2.85~rad/s)(0.300~m)^2}{(0.150~m)^2}$
$\omega_2 = 11.4~rad/s$
(c) We can find the change in the kinetic energy of the block.
$\Delta K = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$
$\Delta K = \frac{1}{2}m(\omega_2~r_2)^2-\frac{1}{2}m(\omega_1~r_1)^2$
$\Delta K = \frac{1}{2}(0.0250~kg)[(11.4~rad/s)(0.150~m)]^2-\frac{1}{2}(0.0250~kg)[(2.85~rad/s)(0.300~m)]^2$
$\Delta K = 0.0274~J$
The kinetic energy of the block increased by 0.0274 J.
(d) The work done in pulling the cord is equal to the change in kinetic energy of the block. The work done in pulling the cord is 0.0274 J.
