A block of mass M is pulled along a horizontal surface by applying a force at angle `theta` with the horizontal. The friction coefficient between the block and the surface is `mu`. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block.

Options

A. `(mumgd)/(cos theta+mu sin theta)`
B. `(mumgd cos theta)/(cos theta+mu sin theta)`
C. `(mumgd sin theta)/(cos theta+mu sin theta)`
D. `(mumgd cos theta)/(cos theta-mu sin theta)`

Explanation

To solve the problem of finding the work done by the applied force when a block of mass \( M \) is pulled along a horizontal surface at an angle \( \theta \) with the horizontal, we can follow these steps:

### Step 1: Identify Forces Acting on the Block
1. The weight of the block acts downward, given by \( W = Mg \).
2. The normal force \( N \) acts upward.
3. The applied force \( F \) has two components:
- Horizontal component: \( F \cos \theta \)
- Vertical component: \( F \sin \theta \)
4. The frictional force \( f \) acts opposite to the direction of motion, given by \( f = \mu N \), where \( \mu \) is the coefficient of friction.

### Step 2: Apply Newton's Second Law in the Vertical Direction
Since the block is moving at a uniform velocity, the net force in the vertical direction must be zero:
\[ N + F \sin \theta = Mg \]
From this, we can express the normal force \( N \):
\[ N = Mg - F \sin \theta \]

### Step 3: Apply Newton's Second Law in the Horizontal Direction
For horizontal motion, since the block moves with uniform velocity, the net force in the horizontal direction is also zero:
\[ F \cos \theta - f = 0 \]
Substituting the frictional force, we have:
\[ F \cos \theta - \mu N = 0 \]
This gives us:
\[ F \cos \theta = \mu N \]

### Step 4: Substitute for Normal Force
Substituting the expression for \( N \) from Step 2 into the equation from Step 3:
\[ F \cos \theta = \mu (Mg - F \sin \theta) \]
Rearranging this equation:
\[ F \cos \theta + \mu F \sin \theta = \mu Mg \]
Factoring out \( F \):
\[ F (\cos \theta + \mu \sin \theta) = \mu Mg \]
Thus, we can solve for \( F \):
\[ F = \frac{\mu Mg}{\cos \theta + \mu \sin \theta} \]

### Step 5: Calculate the Work Done by the Applied Force
The work done \( W \) by the applied force during a displacement \( d \) is given by:
\[ W = F \cdot d \cdot \cos \theta \]
Substituting the expression for \( F \):
\[ W = \left(\frac{\mu Mg}{\cos \theta + \mu \sin \theta}\right) d \cos \theta \]
This simplifies to:
\[ W = \frac{\mu Mg d \cos \theta}{\cos \theta + \mu \sin \theta} \]

### Final Answer
The work done by the applied force during a displacement \( d \) of the block is:
\[ W = \frac{\mu Mg d \cos \theta}{\cos \theta + \mu \sin \theta} \]

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