A block has been placed on an inclined plane with the slope angle `theta`. Block slide down the plane at constant speed. The cofficient of Kinetic friction is equal to

Options

A. `sintheta`
B. `costheta`
C. g
D. `tantheta`

Answer

To solve the problem step by step, we need to analyze the forces acting on the block placed on the inclined plane.

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The gravitational force (weight) acting downwards: \( W = mg \)
2. The normal force (\( R \)) acting perpendicular to the inclined plane.
3. The frictional force (\( F_k \)) acting opposite to the direction of motion (up the incline).

### Step 2: Resolve the gravitational force into components
We can resolve the weight of the block into two components:
- The component parallel to the incline: \( mg \sin \theta \)
- The component perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Write the equation for the forces along the incline
Since the block is sliding down the incline at a constant speed, the net force acting on it is zero. Therefore, we can write:
\[ mg \sin \theta - F_k = 0 \]
This implies:
\[ F_k = mg \sin \theta \]

### Step 4: Express the frictional force in terms of the normal force
The frictional force can also be expressed using the coefficient of kinetic friction (\( \mu_k \)) and the normal force (\( R \)):
\[ F_k = \mu_k R \]

### Step 5: Write the equation for the normal force
The normal force (\( R \)) acting on the block can be expressed as:
\[ R = mg \cos \theta \]

### Step 6: Substitute the expression for the normal force into the frictional force equation
Now, substituting \( R \) into the equation for \( F_k \):
\[ F_k = \mu_k (mg \cos \theta) \]

### Step 7: Set the two expressions for the frictional force equal to each other
Now we have two expressions for \( F_k \):
1. \( F_k = mg \sin \theta \)
2. \( F_k = \mu_k (mg \cos \theta) \)

Setting them equal gives:
\[ mg \sin \theta = \mu_k (mg \cos \theta) \]

### Step 8: Simplify the equation
We can cancel \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \sin \theta = \mu_k \cos \theta \]

### Step 9: Solve for the coefficient of kinetic friction
Now, we can solve for \( \mu_k \):
\[ \mu_k = \frac{\sin \theta}{\cos \theta} \]
\[ \mu_k = \tan \theta \]

### Final Answer
Thus, the coefficient of kinetic friction (\( \mu_k \)) is equal to:
\[ \mu_k = \tan \theta \]

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