A block of mass m = 5 kg is attached to a spring (k = 28 N/m) by a rope that hangs over a pulley of mass M = 7 kg and radius R = 3 cm, as shown in the figure. Treating the pulley as a solid homogeneous disk, neglecting friction at the axle of the pulley, and assuming the system starts from rest with the spring at its natural length, answer the following.

(a) Find the speed of the block after it falls 1 m.


(b) Find the maximum extension of the spring

Someone will gladly critique your work, when shown any.

To find the speed of the block after it falls 1 m, we need to analyze the forces acting on the system of the block and the pulley.

Let's start by considering the block. The force acting on the block is its weight, which is given by:

F_block = m * g

where m is the mass of the block and g is the acceleration due to gravity.

Next, let's consider the pulley. Since the pulley is treated as a solid homogeneous disk, we can treat it as rotating about its center. The force acting on the pulley is the tension in the rope, T. This tension is caused by the weight of the block and the spring force.

The weight of the block creates a clockwise torque on the pulley, while the spring force creates a counterclockwise torque. These torques will cause the pulley to rotate and the block to move.

The torque created by the weight of the block is given by:

τ_weight = (m + M) * g * R

where M is the mass of the pulley and R is its radius.

The torque created by the spring force is given by:

τ_spring = k * x * R

where k is the spring constant and x is the extension or compression of the spring.

Since the system starts from rest and the spring is initially at its natural length, the torque due to the spring force is zero initially. As the block falls, the spring will be compressed, and eventually, the torque due to the spring force will start to oppose the torque due to the weight of the block.

At some point, the torques will be balanced, and the pulley will rotate with a constant angular velocity. When this happens, the block will continue to fall, but its speed will remain constant. This is the point we're interested in, as it will provide us with the speed of the block after it falls 1 m.

To solve for the speed of the block at this point, we need to set the torques equal to each other:

τ_weight = τ_spring

(m + M) * g * R = k * x * R

Now we can solve for the extension or compression of the spring, x:

x = (m + M) * g / k

With the value of x, we can find the maximum extension of the spring by substituting it into the equation for the extension of the spring:

x_max = (m + M) * g / k

Finally, we can calculate the speed of the block after it falls 1 m. Since the block falls vertically downwards, the distance it travels is equal to the extension of the spring:

v = √(2 * g * x)

where g is the acceleration due to gravity and x is the extension of the spring.

By substituting the value of x as calculated earlier, we can find the speed of the block after it falls 1 m.