In the figure below, three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 30.0 kg, mB = 40.0 kg, mC = 28.0 kg.

The image is a horizontal pulley, with mass mA on the left, and B and C on the right side, underneath the pulley and connected by a rope.

(a) When the assembly is released from rest, what is the tension in the cord that connects boxes B and C?
1 N

(b) How far does box A move in the first 0.250 s (assuming it does not reach the pulley)?
2 m

To find the tension in the cord connecting boxes B and C, we need to analyze the forces acting on each box.

First, let's calculate the force of gravitational pull acting on each box:

Fg_A = mA * g
Fg_B = mB * g
Fg_C = mC * g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since the system is in equilibrium (no net acceleration), the tension in the cord connecting B and C will be equal to the difference in gravitational forces on the two boxes:

T_BC = Fg_B - Fg_C

Substituting the values:

T_BC = (mB * g) - (mC * g)
= g * (mB - mC)
= 9.8 * (40 - 28)
= 11.76 N

Therefore, the tension in the cord connecting boxes B and C is 11.76 N.

To find how far box A moves in the first 0.250 s, we can use the formula for displacement under constant acceleration:

d = 0.5 * a * t^2

In this case, box A is not accelerating, so its initial velocity is 0 m/s and the acceleration is also 0 m/s^2. Therefore, the displacement will be 0.

So, box A does not move in the first 0.250 s and the displacement is 0 m.

(a) To find the tension in the cord that connects boxes B and C, we can use Newton's second law of motion. The tension in the cord will be equal to the force required to accelerate the combined masses of B and C.

Since the assembly is released from rest, the net force acting on the system will be the gravitational force acting downwards on the combined masses of B and C. The tension in the cord will be equal in magnitude but opposite in direction to this gravitational force.

The gravitational force acting on the combined masses of B and C can be calculated using the formula:

F = (mB + mC) * g

where mB and mC are the masses of B and C respectively, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Given that mB = 40.0 kg and mC = 28.0 kg, the total mass of B and C is 40.0 kg + 28.0 kg = 68.0 kg.

Plugging these values into the formula, the gravitational force is:

F = (68.0 kg) * (9.8 m/s^2) = 666.4 N

Since the tension in the cord is equal in magnitude but opposite in direction to the gravitational force, the tension is:

Tension = -666.4 N

However, since tension is a scalar quantity, we take the absolute value to get the final answer:

Tension = 666.4 N

So, the tension in the cord that connects boxes B and C is 666.4 N.

(b) To calculate how far box A moves in the first 0.250 s, we can use the kinematic equation:

s = ut + (1/2)at^2

where s is the displacement, u is the initial velocity (which is zero since box A is at rest), a is the acceleration, and t is the time.

Since box A does not reach the pulley, we can assume that its acceleration is the same as the combined acceleration of B and C. The net force acting on the system is the gravitational force, which causes an acceleration of the entire system.

The acceleration of the system can be calculated using Newton's second law:

F = ma

where F is the gravitational force of the system and m is the total mass of the system (mA + mB + mC).

Given that mA = 30.0 kg, mB = 40.0 kg, mC = 28.0 kg, the total mass of the system is:

m = 30.0 kg + 40.0 kg + 28.0 kg = 98.0 kg

Plugging this value into the formula, the acceleration of the system is:

F = (98.0 kg) * (9.8 m/s^2) = 960.4 N

Now, we can substitute the values into the kinematic equation:

s = (0) * (0.250 s) + (1/2) * (960.4 N) * (0.250 s)^2

s = 0 + (1/2) * 960.4 N * 0.0625 s^2

s = 30.025 m

So, box A moves a distance of 30.025 m in the first 0.250 s (assuming it does not reach the pulley).