A block of mass m = 5 kg is attached to a spring (k = 35 N/m) by a rope that hangs over a pulley of mass M = 7 kg and radius R = 4 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

To find the speed of the block after it falls 1 m, we need to consider the conservation of energy in the system. The potential energy of the block decreases as it falls, and this energy is transferred to the kinetic energy of the block.

We can start by calculating the change in potential energy. Since the block falls 1 m, the change in height (Δh) is 1 m. The change in potential energy (ΔPE) is equal to the weight of the block (m * g) multiplied by the change in height (Δh). In this case, the acceleration due to gravity (g) can be assumed to be 9.8 m/s^2.

ΔPE = m * g * Δh
ΔPE = 5 kg * 9.8 m/s^2 * 1 m
ΔPE = 49 J

Next, we can calculate the change in kinetic energy (ΔKE). Since there is no initial kinetic energy (the system starts from rest), the change in kinetic energy is equal to the final kinetic energy of the block (KEf).

ΔKE = KEf

We can equate the change in potential energy to the change in kinetic energy:

ΔPE = ΔKE
49 J = (1/2) * m * v^2

Since the mass of the block (m) is given as 5 kg, we can solve for the final velocity (v).

49 J = (1/2) * 5 kg * v^2
98 J = 5 kg * v^2
v^2 = 98 J / 5 kg
v^2 = 19.6 m^2/s^2
v = √(19.6 m^2/s^2)
v ≈ 4.43 m/s

Therefore, the speed of the block after it falls 1 m is approximately 4.43 m/s.

To find the maximum extension of the spring, we need to consider the equilibrium condition of the system. At maximum extension, the force exerted by the spring is equal in magnitude and opposite in direction to the weight of the block.

The force exerted by the spring is given by Hooke's Law:

F_spring = -k * x

Where k is the spring constant (35 N/m) and x is the displacement from the equilibrium position.

The weight of the block is given by:

F_weight = m * g

Where m is the mass of the block (5 kg) and g is the acceleration due to gravity (9.8 m/s^2).

At maximum extension, the force exerted by the spring is equal in magnitude and opposite in direction to the weight of the block. Therefore, we can equate the two forces:

-k * x = m * g
-35 N/m * x = 5 kg * 9.8 m/s^2
-35 N/m * x = 49 N
x = 49 N / (-35 N/m)
x = -49 N / 35 N/m
x = -1.4 m

Since displacement cannot be negative, we take the absolute value:

x = 1.4 m

Therefore, the maximum extension of the spring is 1.4 m.