A block of mass 3.1 kg hangs vertically from a frictionless pulley of mass 5.9 kg and radius 14 cm. Treat the pulley as a disk. Find:

a) the acceleration of the block;
b) the tension in the rope;
c) the speed of the block after it has fallen 55 cm, assuming it started at rest:

To solve this problem, we will need to apply Newton's second law of motion and the equations for rotational motion. Let's break it down step by step.

a) The acceleration of the block:
We can find the acceleration of the block by considering the forces acting on it. The tension in the rope exerts an upward force, while the gravitational force pulls it downward.

The net force acting on the block is given by:
Net force = Tension - Weight
= Tension - (mass * acceleration due to gravity)

Since the block is in equilibrium along the vertical direction, the net force should be equal to the product of mass and acceleration:
Net force = mass * acceleration

So we have the equation:
Tension - (mass * acceleration due to gravity) = mass * acceleration

The tension in the rope can be related to the rotational motion of the pulley. The torque experienced by the pulley is given by:
Torque = Tension * radius of the pulley

Since the pulley is in equilibrium, the net torque should be zero:
Net torque = Torque due to tension - Torque due to weight = 0

Torque due to tension = Tension * radius
Torque due to weight = (mass of pulley) * (acceleration due to gravity) * (radius)

We can rewrite the equation for net torque as:
Tension * radius - (mass of pulley) * (acceleration due to gravity) * (radius) = 0

Now, we have two equations:
1. Tension - (mass * acceleration due to gravity) = mass * acceleration
2. Tension * radius - (mass of pulley) * (acceleration due to gravity) * (radius) = 0

Solving these two equations simultaneously will give us the values of acceleration and tension.

b) The tension in the rope:
Using the equations above, we can solve for the tension in the rope.

c) The speed of the block after falling 55 cm:
To find the speed of the block, we can use the kinematic equation for motion under constant acceleration:
v^2 = u^2 + 2as

Here, v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration (which we found in part a), and s is the distance fallen (55 cm).

Solving this equation will give us the speed of the block after falling 55 cm.