A skier starts from rest at point A and slides donw the hill without turning or breaking. The friction coefficient is `mu` When he stops at point B, his horizontal displacement is S. whalt is the height difference between points A and B?
(The velocity of the skier is small so that the additional pressure on the snow due to the curvature can vbe neglected. Neglect also the friction of air and the dependence of `mu` on the velocity of the skier)

Options

A. `h=muS`
B. `h=(mu)/(S)`
C. `h=2muS`
D. `h=muS^(2)`

Answer

To solve the problem, we need to determine the height difference \( h \) between points A and B when a skier slides down a hill under the influence of gravity and friction. The skier starts from rest and has a horizontal displacement \( S \) when he stops at point B. The coefficient of friction is \( \mu \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Skier**:
- The gravitational force acting on the skier is \( mg \) (where \( m \) is the mass of the skier and \( g \) is the acceleration due to gravity).
- The frictional force opposing the motion is given by \( F_{\text{friction}} = \mu mg \).

2. **Calculate Work Done by Friction**:
- The work done by the frictional force as the skier moves from A to B can be expressed as:
\[ W_{\text{friction}} = F_{\text{friction}} \cdot d = \mu mg \cdot S \]
- Here, \( S \) is the horizontal displacement.

3. **Relate Work Done to Potential Energy**:
- As the skier descends, the potential energy lost is converted into work done against friction. The potential energy lost when moving down a height \( h \) is given by:
\[ PE_{\text{lost}} = mgh \]
- Setting the work done by friction equal to the potential energy lost gives:
\[ mgh = \mu mg S \]

4. **Cancel Common Terms**:
- We can cancel \( mg \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ h = \mu S \]

5. **Conclusion**:
- The height difference \( h \) between points A and B is:
\[ h = \mu S \]

### Final Answer:
The height difference between points A and B is \( h = \mu S \).