Final answer:
To keep a block static on a frictional inclined plane, the minimum weight required depends on the mass of the block (M2), since both the gravitational component down the ramp and the frictional force resisting it are a function of M2.
Step-by-step explanation:
The question pertains to determining the minimum weight (W1) required to keep a block (M2) stationary on an inclined ramp with a specific coefficient of friction. The answer to this question is (b) Depends on the mass of the block (M2) because to hold the block stationary, we need to balance the force due to gravity along the incline and the force of friction. The gravitational force component acting parallel to the ramp is M2 × g × sin(θ), where g is the acceleration due to gravity (9.81 m/s²) and θ is the inclination angle. The static friction force, which prevents sliding, is given by μ × M2 × g × cos(θ), where μ is the coefficient of friction.
For the block to remain at rest, the required weight (W1) to be applied at the other end of the pulley system must balance the net force along the plane. Since friction is aiding in holding the block by resisting its downward motion and the frictional force depends on the normal force, W1 is essentially adjusting the normal force to increase or decrease friction as needed.
If W1 is at the verge of making the block to move, it can be set to the force of friction which is μ × M2 × g × cos(θ). Then W1= μ × M2 × g × cos(θ) which indicates that the required weight depends on M2, hence providing M2 will allow us to calculate the actual weight needed.