A block is lying on an inclined plane which makes an angle of `60^@` with the horizontal. If coefficient of friction between the block and the plane is 0.25 and g = 10 `ms^(-2)`, the acceleration of block when it is moves along the plane will be

Options

A. `2.5 ms^(-2)`
B. `5 ms^(-2)`
C. `7.25 ms^(-2)`
D. `8.66 ms^(-2)`

Answer

To find the acceleration of a block lying on an inclined plane, we can follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The gravitational force (\(mg\)), which can be resolved into two components:
- Parallel to the incline: \(mg \sin \theta\)
- Perpendicular to the incline: \(mg \cos \theta\)
2. The frictional force (\(f\)), which opposes the motion and is given by \(f = \mu N\), where \(N\) is the normal force.

### Step 2: Calculate the normal force
The normal force \(N\) acting on the block is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 3: Calculate the frictional force
The frictional force can be calculated using the coefficient of friction (\(\mu\)):
\[ f = \mu N = \mu (mg \cos \theta) \]

### Step 4: Write the equation of motion
According to Newton's second law, the net force acting on the block along the incline is equal to the mass times the acceleration (\(ma\)):
\[ mg \sin \theta - f = ma \]
Substituting the expression for frictional force:
\[ mg \sin \theta - \mu (mg \cos \theta) = ma \]

### Step 5: Simplify the equation
We can factor out \(m\) from the equation:
\[ g \sin \theta - \mu g \cos \theta = a \]

### Step 6: Substitute the known values
Given:
- \(\theta = 60^\circ\)
- \(\mu = 0.25\)
- \(g = 10 \, \text{m/s}^2\)

Now, we can calculate \(\sin 60^\circ\) and \(\cos 60^\circ\):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos 60^\circ = \frac{1}{2} \]

Substituting these values into the equation:
\[ a = g \left(\sin 60^\circ - \mu \cos 60^\circ\right) \]
\[ a = 10 \left(\frac{\sqrt{3}}{2} - 0.25 \cdot \frac{1}{2}\right) \]
\[ a = 10 \left(\frac{\sqrt{3}}{2} - \frac{0.25}{2}\right) \]
\[ a = 10 \left(\frac{\sqrt{3}}{2} - \frac{0.125}{1}\right) \]

### Step 7: Calculate the acceleration
Now, we can calculate the numerical value:
\[ a = 10 \left(\frac{\sqrt{3}}{2} - 0.125\right) \]
Using \(\sqrt{3} \approx 1.732\):
\[ a = 10 \left(\frac{1.732}{2} - 0.125\right) \]
\[ a = 10 \left(0.866 - 0.125\right) \]
\[ a = 10 \times 0.741 \]
\[ a \approx 7.41 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the block when it moves along the plane is approximately \(7.41 \, \text{m/s}^2\).

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