Question

A 40 kg skier starts at the top of 12 meter high slope. At the bottom she is traveling 10 m/s how much energy does she lose to friction

Answer 1

To solve this problem we start from the premise that energy is conserved. This means that the energy that the skier has at the beginning of the slope is equal to the energy he has at the end of the slope.

When the skier is at the top of the slope (point 1) his kinetic energy is minimal and his potential energy is maximum. Let's call $E_1$ to the energy in point 1.

$E_1 = mgh$

Where:

mgh = potential energy

When the skier is at the lowest point of the slope (point 2) his kinetic energy is maximum and his potential energy is minimal. Let's call $E_2$ the energy in point 2.

$E_2 = \frac{1}{2}mv^2 + E_{fr}$

Where :

$\frac{1}{2}mv^2$ = kinetic energy

$E_{fr}$ = energy lost by friction

By the law of conservation of energy we know that:

$E_1 = E_2$

So:

$mgh = \frac{1}{2}mv^2 + E_{fr}$

Where:

m = 40kg

v = 10 m/s

h = 12 m

g = 9.8 $m / s^2$

$(40)(9.8)(12) = \frac{1}{2}(40) (10) ^ 2 + E_{fr}$

$E_{fr} = (40)(9.8)(12) - \frac{1}{2}(40)(10)^2$

$E_{fr} = 2704$ J