A particle is pulled a distance `l` up a rough plane inclined at an angle `alpha` to the horizontal by a string inclined at an angle `beta` to the plane `(beta +alpha lt 90^@)`. If the tension the string is T, the normal reaction between the particle and plane is N, the frictional force is F, and the weight of the particle is `w`. write down expression for the work done by each of forces.

Explanation

To solve the problem, we need to determine the work done by each of the forces acting on a particle being pulled up a rough inclined plane. The forces involved are tension (T), normal force (N), frictional force (F), and weight (W) of the particle. Let's break down the solution step by step.

### Step 1: Work Done by the Normal Force (N)
The normal force acts perpendicular to the displacement of the particle along the inclined plane. The work done by a force is given by the formula:

\[ W = F \cdot d \cdot \cos(\theta) \]

where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the displacement.

- Here, the angle \(\theta\) between the normal force \(N\) and the displacement \(l\) is \(90^\circ\).
- Therefore, \(\cos(90^\circ) = 0\).

Thus, the work done by the normal force is:

\[ W_N = N \cdot l \cdot \cos(90^\circ) = 0 \]

### Step 2: Work Done by the Tension Force (T)
The tension in the string acts at an angle \(\beta\) to the inclined plane. The work done by the tension is calculated as follows:

- The angle \(\theta\) between the tension \(T\) and the displacement \(l\) is \(\beta\).

Thus, the work done by the tension is:

\[ W_T = T \cdot l \cdot \cos(\beta) \]

### Step 3: Work Done by the Frictional Force (F)
The frictional force acts opposite to the direction of displacement. The angle between the frictional force and the displacement is \(180^\circ\).

- Therefore, \(\cos(180^\circ) = -1\).

Thus, the work done by the frictional force is:

\[ W_F = F \cdot l \cdot \cos(180^\circ) = -F \cdot l \]

### Step 4: Work Done by the Weight (W)
The weight of the particle acts vertically downward. The angle between the weight and the displacement along the inclined plane is \(90^\circ + \alpha\).

- Therefore, \(\cos(90^\circ + \alpha) = -\sin(\alpha)\).

Thus, the work done by the weight is:

\[ W_W = W \cdot l \cdot \cos(90^\circ + \alpha) = -W \cdot l \cdot \sin(\alpha) \]

### Summary of Work Done by Each Force
1. Work done by the normal force:
\[ W_N = 0 \]
2. Work done by the tension force:
\[ W_T = T \cdot l \cdot \cos(\beta) \]
3. Work done by the frictional force:
\[ W_F = -F \cdot l \]
4. Work done by the weight:
\[ W_W = -W \cdot l \cdot \sin(\alpha) \]