# Efficiency clarification question?

## Athletes who compete in downhill skiing try to lose as little energy as possible. A skier starts from rest at the top of a 65 m hill and skis to the bottom as fast as possible. When she arrives at the bottom, she has a speed of 23 m/s. My question is the following: Why is W_"in" = ∆Eg while ${W}_{\text{out}}$ = ${E}_{k}$? How do you know?

Apr 2, 2017

#### Explanation:

Let $m$ be mass of the skier. As she comes down from a height of $65 m$ to the bottom, the change in her potential energy is given by the equation

$\Delta P E = m g \Delta h = m \times 9.81 \times 65 = m \times 637.65 J$ .......(1)

This is value of ${W}_{\text{in}}$ The operative expression from the question is in bold text in the line below.

as fast as possible $\implies$ Must have maximum speed as she descends the ski slope.

Now kinetic energy which is converted from her original potential energy at the bottom of hill

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times m \times {\left(23\right)}^{2} = m \times 264.5 J$ ......(2)

Law of conservation of energy requires that change in potential energy and kinetic energy gained must be equal. However, from comparison of RHSs of equations (1) and (2), we see that these are not equal.

Evidently, some of potential energy has been lost in the process of skiing down the slope in terms of sliding friction and air friction.

As such this is value of ${W}_{\text{out}}$

Efficiency$= \frac{264.5 m}{637.65 m} \times 100$
=41.48%