A block of mass 4kg rests on an an inclined plane. The inclination of the plane is gradually increased. It is found that when the inclination is 3 in 5 `(sin=theta(3)/(5))` the block just begins to slide down the plane. The coefficient of friction between the block and the plane is.

Options

A. 0.4
B. 0.6
C. 0.8
D. 0.75

Answer

To find the coefficient of friction between the block and the inclined plane, we will follow these steps:

### Step 1: Understanding the Problem
We have a block of mass \( m = 4 \, \text{kg} \) resting on an inclined plane. The inclination angle \( \theta \) is such that \( \sin \theta = \frac{3}{5} \). We need to find the coefficient of friction \( \mu \) when the block just begins to slide down.

### Step 2: Finding the Angle \( \theta \)
Given that \( \sin \theta = \frac{3}{5} \), we can use the Pythagorean theorem to find \( \cos \theta \):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
\[ \cos^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \cos \theta = \frac{4}{5} \]

### Step 3: Finding \( \tan \theta \)
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 4: Relating \( \tan \theta \) to the Coefficient of Friction
The angle \( \theta \) at which the block begins to slide is also the angle of repose. The relationship between the angle of repose and the coefficient of friction is given by:
\[ \tan \theta = \mu \]
Thus, we have:
\[ \mu = \tan \theta = \frac{3}{4} \]

### Step 5: Conclusion
The coefficient of friction \( \mu \) between the block and the inclined plane is:
\[ \mu = 0.75 \]

### Final Answer
The coefficient of friction between the block and the plane is \( 0.75 \).

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