A block of mass m is placed on a smooth wedge of inclination `theta`. The whole system is accelerated horizontally, so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

Options

A. `mgcostheta`
B. `mgsintheta`
C. mg
D. `(mg)/(costheta)`

Answer

To solve the problem, we need to analyze the forces acting on the block placed on the wedge. The wedge is inclined at an angle θ, and the entire system is accelerated horizontally. We will use Newton's laws of motion to derive the required force exerted by the wedge on the block.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The gravitational force acting downward on the block is \( mg \).
- The normal force exerted by the wedge on the block is \( R \), which acts perpendicular to the surface of the wedge.

2. **Resolve Forces into Components:**
- Since the wedge is inclined at an angle θ, we can resolve the normal force \( R \) into two components:
- Horizontal component: \( R \sin \theta \)
- Vertical component: \( R \cos \theta \)

3. **Apply Newton's Second Law in the Horizontal Direction:**
- The block does not slip on the wedge, meaning it accelerates horizontally with the same acceleration \( A \) as the wedge.
- From the horizontal component of the forces, we have:
\[ R \sin \theta = mA \quad \text{(1)} \]

4. **Apply Newton's Second Law in the Vertical Direction:**
- In the vertical direction, the block is in equilibrium because the only forces acting vertically are the weight of the block and the vertical component of the normal force.
- Thus, we have:
\[ R \cos \theta = mg \quad \text{(2)} \]

5. **Solve for the Normal Force \( R \):**
- From equation (2), we can express \( R \) in terms of \( mg \):
\[ R = \frac{mg}{\cos \theta} \]

6. **Conclusion:**
- The force exerted by the wedge on the block is given by:
\[ R = \frac{mg}{\cos \theta} \]

### Final Answer:
The force exerted by the wedge on the block is \( R = \frac{mg}{\cos \theta} \).

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