3)-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.
To find the force that must be exerted in each case, we can use Newton's second law, which states that the force applied to an object is equal to its mass multiplied by its acceleration.
a) To hold the block at rest: In this case, the block is not moving, so its acceleration is zero. Therefore, the force required to hold the block at rest is also zero.
b) To lower it at 2m/s: When lowering the block, we need to consider two forces: the force exerted by the person and the force of gravity acting on the block. The force of gravity can be calculated using the formula F = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The force exerted by the person must be equal to or greater than the force of gravity to lower the block with a constant velocity. So, in this case, the person must exert a force equal to the force of gravity, which is F = 9 kg * 9.8 m/s^2 = 88.2 N.
c) To raise it with an acceleration of 0.5 m/s^2: When raising the block with an acceleration, we again consider two forces: the force exerted by the person and the force of gravity.
To find the force required to raise the block with a given acceleration, we need to consider the net force. The net force is equal to the mass of the block multiplied by the sum of the force of gravity and the force exerted by the person.
Therefore, the equation can be written as:
F_net = m * (g + a)
Substituting the given values, we have:
F_net = 9 kg * (9.8 m/s^2 + 0.5 m/s^2) = 9 kg * 10.3 m/s^2 = 92.7 N.
So, the person must exert a force of 92.7 N to raise the block with an acceleration of 0.5 m/s^2.