## College Physics (4th Edition)

$10.1~J$ of energy has been delivered to the fluid in the beaker.
We can find the initial gravitational potential energy of the block: $U_g = M_b~gh$ $U_g = (0.870~kg)(9.80~m/s^2)(2.50~m)$ $U_g = 21.3~J$ We can find the kinetic energy of the block after it falls 2.5 meters: $KE_b = \frac{1}{2}M_bv^2$ $KE_b = \frac{1}{2}(0.870~kg)(3.00~m/s)^2$ $KE_b = 3.915~J$ We can find the rotational kinetic energy of the pulley: $KE_p = \frac{1}{2}I_p~\omega^2$ $KE_p = \frac{1}{2}M_pR^2~\omega^2$ $KE_p = \frac{1}{2}M_pR^2~(\frac{v}{R})^2$ $KE_p = \frac{1}{2}M_pv^2$ $KE_p = \frac{1}{2}(0.060~kg)(3.00~m/s)^2$ $KE_p = 0.27~J$ We can find the rotational kinetic energy of the spool: $KE_s = \frac{1}{2}I_s~\omega^2$ $KE_s = \frac{1}{2}I_s~(\frac{v}{r})^2$ $KE_s = \frac{1}{2}(0.00140~kg~m^2)~(\frac{3.00~m/s}{0.030~m})^2$ $KE_s = 7.0~J$ We can use conservation of energy to find the energy that $E_f$ has been delivered to the fluid in the beaker: $E_f+KE_b+KE_p+KE_s = U_g$ $E_f = U_g - KE_b - KE_p - KE_s$ $E_f = 21.3~J - 3.915~J - 0.27~J - 7.0~J$ $E_f = 10.1~J$ $10.1~J$ of energy has been delivered to the fluid in the beaker.