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. mgh
B. 2 mgh
C. 4 mgyh
D. `-mgh

Answer

To solve the problem step by step, we can break it down as follows:

### Step 1: Understand the Initial Conditions
The object of mass \( m \) starts from a height \( h \) and slides down the hill. Initially, it possesses gravitational potential energy given by:
\[ PE = mgh \]

### Step 2: Analyze the Motion Down the Hill
As the object slides down, it converts its potential energy into kinetic energy and eventually stops due to friction. The work done against friction is equal to the initial potential energy since the object comes to rest:
\[ \text{Work done by friction} = mgh \]

### Step 3: Determine the Work Done to Return to Initial Position
When the object is to be returned to its initial position, the tangential force must do work against both the frictional force and the gravitational force.

### Step 4: Work Done Against Friction
The work done against friction when returning to the initial position is equal to the energy lost while sliding down, which is:
\[ W_{\text{friction}} = mgh \]

### Step 5: Work Done Against Gravity
When the object is lifted back to the height \( h \), the work done against gravity is also:
\[ W_{\text{gravity}} = mgh \]

### Step 6: Total Work Done by the 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 friction and the work done against gravity:
\[ W_{\text{total}} = W_{\text{friction}} + W_{\text{gravity}} = mgh + mgh = 2mgh \]

### Final Answer
Thus, the work that a tangential force must perform to return the object to its initial position along the same path is:
\[ \boxed{2mgh} \]

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