A block of mass m is lying on an inclined plane. The coefficient of friction between the plane and the block is `mu`. The force `(F_(1))` required to move the block up the inclined plane will be:-

Options

A. `mg sin theta+mumg cos theta`
B. `mg cos theta-mumg sin theta`
C. `mg sin theta-mumg cos theta`
D. `mg cos theta+mumg sin theta`

Explanation

To find the force \( F_1 \) required to move a block of mass \( m \) up an inclined plane with a coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Analyze the Forces Acting on the Block
When the block is on the inclined plane, several forces are acting on it:
- The weight of the block \( mg \) acting vertically downward.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_f \) acting parallel to the inclined plane, opposing the motion.

### Step 2: Resolve the Weight into Components
The weight of the block can be resolved into two components:
- The component parallel to the inclined plane: \( mg \sin \theta \)
- The component perpendicular to the inclined plane: \( mg \cos \theta \)

### Step 3: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

### Step 4: Calculate the Frictional Force
The frictional force \( F_f \) can be calculated using the coefficient of friction \( \mu \):
\[ F_f = \mu N = \mu (mg \cos \theta) \]

### Step 5: Set Up the Equation for the Force Required to Move the Block
To move the block up the inclined plane, the applied force \( F_1 \) must overcome both the gravitational component pulling it down the incline and the frictional force acting against the motion. Therefore, we can express this as:
\[ F_1 = mg \sin \theta + F_f \]

### Step 6: Substitute the Frictional Force into the Equation
Substituting the expression for the frictional force into the equation gives:
\[ F_1 = mg \sin \theta + \mu (mg \cos \theta) \]

### Final Expression
Thus, the force required to move the block up the inclined plane is:
\[ F_1 = mg \sin \theta + \mu mg \cos \theta \]