A body is moving is down an inclined plane of slope `37^@` the coefficient of friction between the body and the plane varies as `mu=0.3x`, where x is the distance traveled down the plane by the body. The body will have maximum speed. `(sin 37^@ =(3)/(5))` . a) at x = 1.16 m b) at x = 2m c) at bottom most point of the plane d) at x = 2.5 m

Options

A. at `x= 1.16 m`
B. at `x= 2m`
C. at bottommost point of the plane
D. at `x = 2.5 m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Analyze the forces acting on the body
The body is moving down an inclined plane with an angle of \(37^\circ\). The forces acting on the body include:
- The gravitational force \(mg\) acting downwards.
- The component of gravitational force acting down the incline: \(mg \sin \theta\).
- The normal force \(N\) acting perpendicular to the incline: \(N = mg \cos \theta\).
- The frictional force \(F_k\) acting up the incline, which is given by \(F_k = \mu N\).

### Step 2: Write the expression for the net force
The net force acting on the body can be expressed as:
\[ F_{\text{net}} = mg \sin \theta - F_k \]
Substituting the expression for frictional force:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
Given that \(\mu = 0.3x\), we can rewrite the equation as:
\[ F_{\text{net}} = mg \sin \theta - 0.3x \cdot mg \cos \theta \]

### Step 3: Set the net force to zero for maximum speed
At maximum speed, the net force will be zero:
\[ 0 = mg \sin \theta - 0.3x \cdot mg \cos \theta \]
This simplifies to:
\[ mg \sin \theta = 0.3x \cdot mg \cos \theta \]
Dividing both sides by \(mg\) (assuming \(mg \neq 0\)):
\[ \sin \theta = 0.3x \cos \theta \]

### Step 4: Solve for \(x\)
Rearranging the equation gives:
\[ x = \frac{\sin \theta}{0.3 \cos \theta} \]
Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\[ x = \frac{\tan \theta}{0.3} \]

### Step 5: Calculate \(\tan \theta\)
Given that \(\sin 37^\circ = \frac{3}{5}\), we can calculate \(\cos 37^\circ\):
\[ \cos 37^\circ = \sqrt{1 - \sin^2 37^\circ} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Now, we can find \(\tan 37^\circ\):
\[ \tan 37^\circ = \frac{\sin 37^\circ}{\cos 37^\circ} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 6: Substitute \(\tan \theta\) into the equation for \(x\)
Substituting \(\tan 37^\circ\) into the equation for \(x\):
\[ x = \frac{\frac{3}{4}}{0.3} = \frac{3}{4 \cdot 0.3} = \frac{3}{1.2} = 2.5 \text{ m} \]

### Conclusion
The body will have maximum speed at a distance of \(x = 2.5 \text{ m}\).

### Final Answer
The correct option is **d) at x = 2.5 m**.

---