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 in the previous problem, we replace the man by a block of mass m and release it from top of the inclined plane, and let it gain a speed v, then

Options

A. (a) `W_(f riction)=-mgh+1/2mv^2`
B. (b) `W_(gravity)=-mgh`
C. (c) `W_(f riction)=0`
D. (d) `W_(f riction)=-mumgx`, where x is the horizontal distance covered and `mu` is the coefficient of friction between the block and the ground.

Explanation

To solve the problem step by step, we will analyze the situation of a block sliding down an inclined plane and apply the work-energy theorem.

### Step 1: Understand the Problem
We have a block of mass `m` that is released from the top of an inclined plane and slides down, gaining a speed `v`. The height raised is `h`. We need to analyze the work done by gravity, friction, and the work done by the block.

### Step 2: Identify Forces Acting on the Block
The forces acting on the block are:
- Weight (mg) acting downwards.
- Normal force (N) acting perpendicular to the surface of the incline.
- Frictional force (f) acting opposite to the direction of motion.

### Step 3: Apply the Work-Energy Theorem
According to the work-energy theorem:
\[ W_{\text{total}} = \Delta KE \]
Where:
- \( W_{\text{total}} \) is the work done by all forces.
- \( \Delta KE \) is the change in kinetic energy.

The change in kinetic energy when the block goes from rest to speed `v` is:
\[ \Delta KE = \frac{1}{2} mv^2 - 0 = \frac{1}{2} mv^2 \]

### Step 4: Calculate Work Done by Each Force
1. **Work Done by Gravity (W_gravity)**:
The work done by gravity as the block moves down the incline is given by:
\[ W_{\text{gravity}} = mgh \]
This is positive because the displacement is in the direction of the gravitational force.

2. **Work Done by Normal Force (W_normal)**:
The normal force acts perpendicular to the direction of motion, so:
\[ W_{\text{normal}} = 0 \]

3. **Work Done by Friction (W_friction)**:
The work done by friction is negative because it opposes the motion:
\[ W_{\text{friction}} = -f \cdot d \]
Where \( d \) is the distance traveled along the incline. The frictional force can be expressed as:
\[ f = \mu N = \mu mg \cos(\theta) \]
Thus, the work done by friction becomes:
\[ W_{\text{friction}} = -\mu mg \cos(\theta) \cdot d \]
If we express \( d \) in terms of the horizontal distance \( x \) traveled, we can rewrite this as:
\[ W_{\text{friction}} = -\mu mgx \]

### Step 5: Combine the Work Done
Now we can combine the work done by all forces:
\[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{normal}} + W_{\text{friction}} \]
Substituting the values:
\[ W_{\text{total}} = mgh + 0 - \mu mgx \]
This gives us:
\[ W_{\text{total}} = mgh - \mu mgx \]

### Step 6: Set the Total Work Equal to Change in Kinetic Energy
Setting the total work equal to the change in kinetic energy:
\[ mgh - \mu mgx = \frac{1}{2} mv^2 \]

### Step 7: Analyze the Options
Now, we can analyze the options provided in the question:
1. **Option A**: Work done by friction will be \(-mgh + \frac{1}{2} mv^2\) - This is correct.
2. **Option B**: Work done by gravity will be \(-mgh\) - This is incorrect as it should be \(+mgh\).
3. **Option C**: Work done by friction is zero - This is incorrect as we calculated it to be non-zero.
4. **Option D**: Work done by friction is \(-\mu mgx\) - This is correct.

### Conclusion
The correct options are A and D.

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