Answer
(a) The speed at the bottom of the slope is 30.4 m/s.
(b) The speed after crossing the patch of soft snow is 12.9 m/s.
(c) The magnitude of the average force exerted on the skier by the snowdrift is $2.00\times 10^3~N$ and it is directed opposite to the skier's direction of motion.
Work Step by Step
(a) $K_2+U_2 = K_1+U_1+W$
$\frac{1}{2}mv_2^2+0=0+mgh - 10,500~J$
$v_2^2=\frac{2mgh - (2)(10,500~J)}{m}$
$v_2=\sqrt{\frac{2mgh - (2)(10,500~J)}{m}}$
$v_2=\sqrt{\frac{(2)(60.0~kg)(9.80~m/s^2)(65.0~m) - (2)(10,500~J)}{60.0~kg}}$
$v_2 = 30.4~m/s$
(b) $K_3+U_3 = K_2+U_2+W$
$\frac{1}{2}mv_3^2+0 = \frac{1}{2}mv_2^2+0-F_f~d-(160~N)~d$
$v_3^2 = \frac{mv_2^2-2mg~\mu_k~d-(2)(160~N)~d}{m}$
$v_3 = \sqrt{\frac{mv_2^2-2mg~\mu_k~d-(2)(160~N)~d}{m}}$
$v_3 = \sqrt{\frac{(60.0~kg)(30.4~m/s)^2-(2)(60.0~kg)(9.80~m/s^2)(0.20)(82.0~m)-(2)(160~N)(82.0~m)}{60.0~kg}}$
$v_3 = 12.9~m/s$
The speed after crossing the patch is 12.9 m/s.
(c) $F~d = K_3$
$F = \frac{mv_3^2}{2d}$
$F = \frac{(60.0~kg)(12.9~m/s)^2}{(2)(2.5~m)}$
$F = 2.00\times 10^3~N$
The magnitude of the average force exerted on the skier is $2.00\times 10^3~N$ and it is directed opposite to the skier's direction of motion.
