A 58.0 kg skier is moving at 6.00 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.65 m long. The coefficient of kinetic friction between this patch and her skis is 0.310. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.50 m high.

Required:
a. How fast is the skier moving when she gets to the bottom of the hill?
b. How much internal energy was generated in crossing the rough patch?

a) To find the speed of the skier you take into account that, the work done by the friction surface on the skier is equal to the change in the kinetic energy:

[tex]-W_f=\Delta K=\frac{1}{2}m(v^2-v_o^2)\\\\-F_fd=\frac{1}{2}m(v^2-v_o^2)[/tex]

(the minus sign is due to the work is against the motion of the skier)

m: mass of the skier = 58.0 kg

v: final speed = ?

vo: initial speed = 6.00 m/s

d: distance traveled by the skier in the rough patch = 3.65 m

Ff: friction force = Mgμ

g: gravitational acceleration = 9.8 m/s^2

μ: friction coefficient = 0.310

You solve the equation (1) for v:

[tex]v=\sqrt{\frac{2F_fd}{m}+v_o^2}=\sqrt{\frac{2mg\mu d}{m}+v_o^2}\\\\v=\sqrt{-2g\mu d+v_o^2}[/tex]

Next, you replace the values of all parameters:

[tex]v=\sqrt{-2(9.8m/s^2)(0.310)(3.65m)+(6.00m/s)^2}=3.71\frac{m}{s}[/tex]

The speed after the skier has crossed the roug path is 3.71m/s

b) The work done by the rough patch is the internal energy generated:

[tex]U=W_fd=F_fd=mg\mu d\\\\U=(58.0kg)(9.8m/s^2)(0.310)(3.50m)=616.71\ J[/tex]

The internal energy generated is 616.71J