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 of finding the force exerted by the wedge on the block, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The block of mass \( m \) experiences its weight \( mg \) acting downwards.
- Since the wedge is inclined at an angle \( \theta \), the weight can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

2. **Consider the Acceleration of the System:**
- The entire system (wedge and block) is accelerating horizontally with an acceleration \( A \).
- Due to this horizontal acceleration, a pseudo force \( F_{\text{pseudo}} = ma \) acts on the block in the opposite direction of the acceleration of the wedge.

3. **Resolve the Pseudo Force:**
- The pseudo force can also be resolved into components:
- Perpendicular to the incline: \( ma \cos \theta \)
- Parallel to the incline: \( ma \sin \theta \)

4. **Set Up the Equilibrium Conditions:**
- For the block to not slip on the wedge, the net force in the direction perpendicular to the incline must be zero.
- Thus, we can write the equation:
\[ N - mg \cos \theta - ma \sin \theta = 0 \]
- Where \( N \) is the normal force exerted by the wedge on the block.

5. **Solve for Normal Force \( N \):**
- Rearranging the equation gives:
\[ N = mg \cos \theta + ma \sin \theta \]

6. **Substitute the Expression for Acceleration \( A \):**
- From the equilibrium of forces along the incline, we have:
\[ ma \cos \theta = mg \sin \theta \]
- This simplifies to:
\[ a = g \tan \theta \]
- Substitute \( a \) back into the equation for \( N \):
\[ N = mg \cos \theta + m(g \tan \theta) \sin \theta \]

7. **Simplify the Expression:**
- Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ N = mg \cos \theta + mg \sin^2 \theta \]
- Factor out \( mg \):
\[ N = mg \left( \cos \theta + \sin^2 \theta \right) \]

### Final Answer:
The force exerted by the wedge on the block is:
\[ N = mg \left( \cos \theta + \sin^2 \theta \right) \]