A 9-kg block is held by a pulley system shown in fig. What force must the person exert in the following cases: a) to hold the block at rest. b) to lower it at 2m/s. c) to raise it with an acceleration of 0.5 m/s2? Ignore the mass of small pulley.

Answer Thi

Ummmm.

There seems to be a distinct lack of a figure...

To answer these questions, we need to analyze the forces acting on the block in each case.

a) To hold the block at rest:
In this case, the block is not moving, so the net force acting on it must be zero. The person needs to exert a force equal to the weight of the block to balance out the force of gravity pulling the block down. The weight of the block can be calculated using the equation:

Weight = mass * gravitational acceleration

W = 9 kg * 9.8 m/s^2 = 88.2 N

Therefore, the person must exert a force of 88.2 N to hold the block at rest.

b) To lower it at 2 m/s:
When the block is being lowered, the force exerted by the person needs to counteract not only the weight of the block but also the downward acceleration. The net force can be calculated using the equation:

Net force = mass * (acceleration due to gravity + downward acceleration)

In this case, the downward acceleration is given as 2 m/s^2.

Net force = 9 kg * (9.8 m/s^2 + 2 m/s^2) = 127.8 N

Therefore, the person must exert a force of 127.8 N to lower the block at 2 m/s.

c) To raise it with an acceleration of 0.5 m/s^2:
When the block is being raised, the force exerted by the person needs to counteract not only the weight of the block but also the upward acceleration. The net force can be calculated using the equation:

Net force = mass * (acceleration due to gravity - upward acceleration)

In this case, the upward acceleration is given as 0.5 m/s^2.

Net force = 9 kg * (9.8 m/s^2 - 0.5 m/s^2) = 85.5 N

Therefore, the person must exert a force of 85.5 N to raise the block with an acceleration of 0.5 m/s^2.