An object of mass `m` slides down a hill of arbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended is `h`. The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangerial force must perform to return the object to its initial position along the same path is

Options

A. (a) `mgh`
B. (b) `-mgh`
C. (c) `-2mgh`
D. (d) `2mgh`

Explanation

To solve the problem, we need to determine the total work done by a tangential force to return the object to its initial position along the same path after it has slid down a hill and stopped due to friction.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- An object of mass \( m \) slides down a hill and descends a vertical height \( h \).
- The object stops due to friction, which varies along the path.
- We need to find the work done by a tangential force to bring the object back to its starting point.

2. **Work Against Gravity**:
- When the object is raised back to its initial position, it must overcome the gravitational potential energy lost while sliding down.
- The work done against gravity (to raise the object back up) is given by:
\[ W_{\text{gravity}} = mgh \]

3. **Work Done Against Friction**:
- As the object slides down, friction does negative work on the object, which can be equated to the loss of potential energy.
- The total work done by friction as the object slides down is equal to the potential energy lost:
\[ W_{\text{friction}} = -mgh \]
- Since the work done by friction is negative, the work done against friction when bringing the object back up will be the same magnitude but positive:
\[ W_{\text{against friction}} = mgh \]

4. **Total Work Done by Tangential Force**:
- The total work done by the tangential force to return the object to its initial position is the sum of the work done against gravity and the work done against friction:
\[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{against friction}} = mgh + mgh = 2mgh \]

5. **Final Answer**:
- Therefore, the total work that a tangential force must perform to return the object to its initial position along the same path is:
\[ \boxed{2mgh} \]