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 = 24.0 kg, mB = 40.0 kg, mC = 30.0 kg.

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

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

So how can we state the appropriate answer to part (a)...?

To solve this problem, we need to apply the principles of Newton's second law and the concept of tension in a system of connected objects.

(a) To find the tension in the cord connecting boxes B and C, we need to analyze the forces acting on each box. Let's assume the positive direction is upward.

For Box B:
- The weight of Box B is given by (mB * g), where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The tension in the cord connected to Box B is TBC (the tension we are trying to find).
- Since Box B is accelerating upward, the net force acting on it is given by (mB * a), where a is its acceleration. Therefore, the net force is (mB * a).

For Box C:
- The weight of Box C is given by (mC * g).
- The tension in the cord connected to Box C is TBC.
- Since Box C is also accelerating upward, the net force acting on it is given by (mC * a).

Since the cord connecting Box B and C is connected over a pulley with negligible friction, the tension in this cord is the same. Therefore, TBC is the tension in the cord connecting Boxes B and C.

Now, let's apply Newton's second law in the vertical direction for Boxes B and C:

For Box B: TBC - (mB * g) = (mB * a)
For Box C: TBC - (mC * g) = (mC * a)

We can solve these two equations simultaneously to find TBC. Substituting the given values into the equations:

For Box B: TBC - (40.0 kg * 9.8 m/s^2) = (40.0 kg * a)
For Box C: TBC - (30.0 kg * 9.8 m/s^2) = (30.0 kg * a)

Now, solve these equations simultaneously to find TBC.

(b) To find how far Box A moves in the first 0.250 s, we can apply the equations of motion for constant acceleration. Since Box A does not reach the pulley, it moves toward Box B.

The equation of motion we can use is:
s = ut + (1/2)at^2

Assuming Box A starts from rest (u = 0), we can simplify the equation to:
s = (1/2)at^2

Substituting the given values into the equation:
s = (1/2) * a * (0.250 s)^2

Now, calculate the distance Box A moves using the acceleration value obtained in part (a).

Remember to always double-check units and ensure that the final answer is in meters.

Note: Make sure to check if the values and assumptions provided in the problem are correct, and if any additional information is given. If so, include those in the calculations.

the answer to (a) is T2

(a) m(A)a=T1

m(B)a=m(B)g-T1+T2
m(C)a=m(C)g-T2
a{m(A) + m(B)+m(C)} = T1+ m(B)g-T1+T2 + m(C)g-T2 =g{ m(B)+m(C)}
a= g{ m(B)+m(C)}/{ m(A) + m(B)+m(C)} =9.8(40+30)/(24+40+30) =7.3 m/s²


T1 =m(A)a=24•7.3=175.2 N
T2 = m(C)(g-a) = 30•(9.8-7.3)= 75 N

(b) s=at²/2 = 7.3•0.25²/2 = 0.228 m