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"`
Which of the following options is correct regarding the various work done?

Options

A. (a) `W_(gravity)=-mgh`
B. (b) `W_(f riction)gt0`
C. (c) `W_(man)=mgh+1/2mv^2`
D. (d) `W_(f riction)=0`

Explanation

To solve the problem, we will analyze the work done by gravity, friction, and the man while he runs up the inclined plane.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man**:
- The forces acting on the man include:
- Gravitational force (downward): \( F_g = mg \)
- Normal force (perpendicular to the incline): \( F_n \)
- Frictional force (opposing the motion): \( F_f \)

2. **Apply Work-Energy Theorem**:
- According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy:
\[ W_{\text{total}} = W_{\text{man}} + W_{\text{gravity}} + W_{\text{normal}} + W_{\text{friction}} = \Delta KE \]
- The change in kinetic energy when the man speeds up from rest to speed \( v \) is:
\[ \Delta KE = \frac{1}{2} mv^2 - 0 = \frac{1}{2} mv^2 \]

3. **Calculate Work Done by Gravity**:
- The work done by gravity when the man rises through height \( h \) is:
\[ W_{\text{gravity}} = -mgh \]
- The negative sign indicates that the work done by gravity is in the opposite direction to the displacement.

4. **Calculate Work Done by Normal Force**:
- The work done by the normal force is zero because the normal force acts perpendicular to the direction of motion:
\[ W_{\text{normal}} = 0 \]

5. **Calculate Work Done by Friction**:
- The work done by friction is also zero. This is because, although friction acts on the man, there is no displacement in the direction of the frictional force when the man is running:
\[ W_{\text{friction}} = 0 \]

6. **Set Up the Equation**:
- Now, substituting the values into the work-energy theorem:
\[ W_{\text{man}} + (-mgh) + 0 + 0 = \frac{1}{2} mv^2 \]
- Rearranging gives:
\[ W_{\text{man}} - mgh = \frac{1}{2} mv^2 \]
- Therefore, the work done by the man is:
\[ W_{\text{man}} = mgh + \frac{1}{2} mv^2 \]

7. **Conclusion**:
- The correct statements regarding the work done are:
- \( W_{\text{gravity}} = -mgh \) (Correct)
- \( W_{\text{friction}} = 0 \) (Correct)
- \( W_{\text{normal}} = 0 \) (Correct)
- \( W_{\text{man}} = mgh + \frac{1}{2} mv^2 \) (Correct)

### Final Answer:
The correct options regarding the various work done are A, C, and D.