A man of mass `m` speeds up while running from rest to a speed v in a straight track along an inclined plane, after raising through a height `h`.
`W_(gravity)="work done by gravity on the man"`
`W_(friction)="work done by friction on the man"`
`W_(man)="work done by man"`
If instead of moving up the plane, the man increases his speed to the value v while moving down the inclined plane through the same vertical distance h, then

Options

A. (a) `W_(f riction)gt0`
B. (b) `W_(f riction)=-mgh+1/2mv^2`
C. (c) Work done by the man can be positive, negative or zero
D. (d) `W_(f riction)+W_(man)=-mgh+1/2mv^2`

Explanation

To solve the problem step-by-step, we will analyze the situation where a man of mass \( m \) runs up and down an inclined plane, changing his speed and the work done by various forces.

### Step 1: Understand the scenario
The man starts from rest and accelerates to a speed \( v \) while moving up an inclined plane through a height \( h \). We need to analyze the work done by gravity, friction, and the man in both scenarios: moving up and moving down the incline.

### Step 2: Work-Energy Theorem
According to the work-energy theorem, the work done by all forces acting on an object is equal to the change in its kinetic energy:
\[ W_{\text{total}} = \Delta KE \]
Where:
- \( \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2} mv^2 - 0 = \frac{1}{2} mv^2 \)

### Step 3: Forces acting on the man
When the man is running up the incline:
- Weight \( W = mg \) acts downward.
- Normal force \( N \) acts perpendicular to the incline.
- Friction force \( f \) acts opposite to the direction of motion.

### Step 4: Work done by each force
1. **Work done by gravity**:
The work done by gravity when moving up is:
\[ W_{\text{gravity}} = -mgh \]
(negative because it acts downward while the man moves upward)

2. **Work done by normal force**:
The normal force does no work since it acts perpendicular to the displacement:
\[ W_{\text{normal}} = 0 \]

3. **Work done by friction**:
Assuming friction is present, but since the foot is on the ground, the displacement in the direction of friction is zero:
\[ W_{\text{friction}} = 0 \]

4. **Work done by the man**:
Let \( W_{\text{man}} \) be the work done by the man. According to the work-energy theorem:
\[ W_{\text{man}} + W_{\text{gravity}} + W_{\text{normal}} + W_{\text{friction}} = \Delta KE \]
Plugging in the values:
\[ W_{\text{man}} - mgh + 0 + 0 = \frac{1}{2} mv^2 \]
Thus, we have:
\[ W_{\text{man}} = mgh + \frac{1}{2} mv^2 \]

### Step 5: Analyzing the downward motion
Now, if the man runs down the incline while increasing his speed to \( v \) through the same height \( h \):
1. The work done by gravity is now positive:
\[ W_{\text{gravity}} = mgh \]

2. The work done by the normal force and friction remains the same:
\[ W_{\text{normal}} = 0, \quad W_{\text{friction}} = 0 \]

3. Applying the work-energy theorem again:
\[ W_{\text{man}} + mgh + 0 + 0 = \frac{1}{2} mv^2 \]
Rearranging gives:
\[ W_{\text{man}} = \frac{1}{2} mv^2 - mgh \]

### Conclusion
From the two scenarios, we can summarize:
- When moving up: \( W_{\text{man}} = mgh + \frac{1}{2} mv^2 \)
- When moving down: \( W_{\text{man}} = \frac{1}{2} mv^2 - mgh \)