Question

A 62.0-kg skier is moving at 6.10 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.10 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 mm high.(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

b) 747.3 J

a) 7.88 m/s

Explanation:

b)

We start by solving part b) first.

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

$F_f=\mu mg$

where:

$\mu=0.300$ is the coefficient of friction

m = 62.0 kg is the mass of the skier

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the work done by friction is:

$W=-F_f d =-\mu mg d$

where

d = 4.10 m is the length of the rough patch

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

$W=-(0.300)(62.0)(9.8)(4.10)=-747.3 J$

So, the internal energy generated in crossing the rough patch is 747.3 J.

a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

$E=K_i +U_i = \frac{1}{2}mu^2 + mgh$

where

m = 62.0 kg is the mass

u = 6.10 m/s is the initial speed

h = 2.50 m is the height of the hill

After crossing the rough patch, the new mechanical energy is

$E'=E+W$

where

W = -747.3 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

$E' = K_f = \frac{1}{2}mv^2$

where v is the final speed.

According to the law of conservation of energy, we can write:

$E+W=E'$

So we find v:

$\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\v=\sqrt{u^2+2gh+\frac{2W}{m}}=\sqrt{6.10^2+2(9.8)(2.50)+\frac{2(-747.3)}{62.0}}=7.88 m/s$