A 200 N block rests on a frictionless inclined plane of angle of inclination $30^{\circ}$. A cord attached to the block passes over a frictionless pulley at the top of the plane is attached to a second block. If the system is in equilibrium, then the weight of second block (in N) approximately is:

Question

Accepted Answer

To determine the weight of the second block that keeps the system in equilibrium, we need to analyze the forces acting on the 200 N block on the inclined plane.

1. **Forces on the 200 N Block:**
   - The gravitational force acting on the block is 200 N, directed vertically downward.
   - This force can be resolved into two components:
     - A component parallel to the inclined plane: $ F_{\text{parallel}} = 200 \sin(30^\circ) $
     - A component perpendicular to the inclined plane: $ F_{\text{perpendicular}} = 200 \cos(30^\circ) $

Since the plane is frictionless, only the parallel component affects the equilibrium condition.

2. **Calculating the Parallel Component:**
   $$
   F_{\text{parallel}} = 200 \sin(30^\circ) = 200 \times 0.5 = 100 \, \text{N}
   $$

3. **Equilibrium Condition:**
   For the system to be in equilibrium, the tension in the cord (which is equal to the weight of the second block) must balance the parallel component of the gravitational force on the 200 N block. Therefore, the weight of the second block must be equal to $ F_{\text{parallel}} $.

4. **Weight of the Second Block:**
   The weight of the second block is 100 N, which matches Option 1.

5. **Explanation for Other Options:**
   - **Option 2 (86.6 N):** This value might be derived from a miscalculation involving the cosine component, which is not relevant for the equilibrium condition on a frictionless plane.
   - **Option 3 (173 N):** This value does not correspond to any logical calculation based on the given scenario. It is significantly higher than the required tension to balance the parallel component.
   - **Option 4 (115.5 N):** This value is also incorrect as it does not match the calculated parallel component of 100 N.

In conclusion, the correct weight of the second block that maintains equilibrium is 100 N, as given in Option 1. This is supported by the calculation of the parallel component of the gravitational force acting on the 200 N block on the inclined plane.