A wooden block of mass M resting on a rough horizontal surface is pulled with a force T at an angle q to the horizontal. If m is coefficient of kinetic friction between the block and the surface, the acceleration of the block is

Options

A. `(T cos theta)/(M) - mu g`
B. `T/m (cos theta - sin theta) - mu g`
C. `T/m (cos theta - mu sin theta) - mu g`
D. `(mu T cos theta)/(M) - mu g`

Answer

To find the acceleration of a wooden block of mass \( M \) being pulled with a force \( T \) at an angle \( \theta \) to the horizontal on a rough surface, we can follow these steps:

### Step 1: Analyze the Forces Acting on the Block
The forces acting on the block include:
- The gravitational force \( W = Mg \) acting downward.
- The normal force \( N \) acting upward.
- The pulling force \( T \) at an angle \( \theta \) which can be resolved into two components:
- Horizontal component: \( T \cos \theta \)
- Vertical component: \( T \sin \theta \)
- The frictional force \( f \) acting opposite to the direction of motion.

### Step 2: Write the Equation for the Vertical Forces
In the vertical direction, the forces must balance. Thus, we have:
\[ N + T \sin \theta = Mg \]
From this equation, we can express the normal force \( N \):
\[ N = Mg - T \sin \theta \]

### Step 3: Write the Equation for the Frictional Force
The frictional force \( f \) can be expressed as:
\[ f = \mu N = \mu (Mg - T \sin \theta) \]
where \( \mu \) is the coefficient of kinetic friction.

### Step 4: Write the Equation for the Horizontal Forces
In the horizontal direction, the net force acting on the block is given by the difference between the horizontal component of the pulling force and the frictional force:
\[ T \cos \theta - f = Ma \]
Substituting the expression for \( f \):
\[ T \cos \theta - \mu (Mg - T \sin \theta) = Ma \]

### Step 5: Rearranging the Equation
Now, rearranging the equation gives:
\[ T \cos \theta - \mu Mg + \mu T \sin \theta = Ma \]
Combining the terms involving \( T \):
\[ (T \cos \theta + \mu T \sin \theta) - \mu Mg = Ma \]

### Step 6: Solve for Acceleration \( a \)
Now, we can isolate \( a \):
\[ a = \frac{T \cos \theta + \mu T \sin \theta - \mu Mg}{M} \]

### Final Expression for Acceleration
Thus, the acceleration \( a \) of the block is given by:
\[ a = \frac{T \cos \theta + \mu T \sin \theta - \mu Mg}{M} \]