Question

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?

Answer 1

a) v = 3.71m/s

b) U = 616.71 J

Step-by-step explanation:

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:

`-W_f=\Delta K=(1)/(2)m(v^2-v_o^2)\\\\-F_fd=(1)/(2)m(v^2-v_o^2)`

(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:

`v=\sqrt{(2F_fd)/(m)+v_o^2}=\sqrt{(2mg\mu d)/(m)+v_o^2}\\\\v=√(-2g\mu d+v_o^2)`

Next, you replace the values of all parameters:

`v=√(-2(9.8m/s^2)(0.310)(3.65m)+(6.00m/s)^2)=3.71(m)/(s)`

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:

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

The internal energy generated is 616.71J