A block of weight 200N is pulled along a rough horizontal surface at constant speed by a force 100N acting at an angle `30^(@)` above the horizontal. The coefficient of friction between the block and the surface is

Options

A. 0.43
B. 0.58
C. 0.75
D. 0.85

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the block
1. The weight of the block (W) = 200 N (acting downwards).
2. The pulling force (F) = 100 N at an angle of 30° above the horizontal.
3. The normal force (N) acting perpendicular to the surface.
4. The frictional force (f_r) opposing the motion.

### Step 2: Resolve the pulling force into components
- The horizontal component of the pulling force (F_x) is given by:
\[ F_x = F \cos(\theta) = 100 \cos(30°) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \, \text{N} \]

- The vertical component of the pulling force (F_y) is given by:
\[ F_y = F \sin(\theta) = 100 \sin(30°) = 100 \times \frac{1}{2} = 50 \, \text{N} \]

### Step 3: Apply Newton's second law in the vertical direction
Since the block is moving at constant speed, the net force in the vertical direction is zero. Thus, we can write:
\[ N + F_y = W \]
Substituting the known values:
\[ N + 50 = 200 \]
Solving for N:
\[ N = 200 - 50 = 150 \, \text{N} \]

### Step 4: Apply Newton's second law in the horizontal direction
Since the block is moving at constant speed, the net force in the horizontal direction is also zero. Thus, we can write:
\[ f_r = F_x \]
Substituting the expression for frictional force:
\[ f_r = \mu N \]
Where \(\mu\) is the coefficient of friction. Therefore:
\[ \mu N = F_x \]
Substituting the values we have:
\[ \mu \cdot 150 = 50\sqrt{3} \]

### Step 5: Solve for the coefficient of friction (\(\mu\))
Rearranging the equation gives:
\[ \mu = \frac{50\sqrt{3}}{150} \]
Simplifying:
\[ \mu = \frac{\sqrt{3}}{3} \]

### Step 6: Calculate the numerical value
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ \mu \approx \frac{1.732}{3} \approx 0.577 \]

Thus, the coefficient of friction is approximately \(0.58\).

### Final Answer
The coefficient of friction between the block and the surface is approximately **0.58**.

---