## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$v = 2.4~m/s$
The sum of the work done by the tension and the friction will be equal to the kinetic energy. We can find the work done by the tension. $W_T = T~d~cos(\theta)$ $W_T = (30~N)(3.0~m)~cos(30^{\circ})$ $W_T = 77.94~J$ We can find the work done by friction. $W_f = F_f~d~cos(180^{\circ})$ $W_f = F_N~\mu_k~d~cos(180^{\circ})$ $W_f = [mg-T~sin(30^{\circ})]~\mu_k~d~cos(180^{\circ})$ $W_f = [(10~kg)(9.80~m/s^2)-(30~N)~sin(30^{\circ})](0.20)(3.0~m)~cos(180^{\circ})$ $W_f = -49.8~J$ We can find Paul's speed. $KE = W_T+W_f$ $\frac{1}{2}mv^2 = W_T+W_f$ $v^2 = \frac{2(W_T+W_f)}{m}$ $v = \sqrt{\frac{2(W_T+W_f)}{m}}$ $v = \sqrt{\frac{(2)(77.94~J-49.8~J)}{10~kg}}$ $v = 2.4~m/s$