In the figure below, Block A has mass 2.14 kg and Block B has mass 6.85 kg. Bloc
ID: 2095109 • Letter: I
Question
In the figure below, Block A has mass 2.14 kg and Block B has mass 6.85 kg. Block A slides on the horizontal table with coefficient of kinetic friction 0.331. The pulley spins with the taut string, so that the string does not slip as it passes over the pulley (what is the relationship between string speed and tangential speed of the outer surface of the pulley?). Treat the pulley as a solid disc with mass 0.421 kg and radius 1 cm. What is the speed of Block B after it has fallen by 1.03 m? Assume that both blocks are initially at rest and that the pulley spins without losing energy to friction.
Explanation / Answer
First, the string speed and the tangential speed of the outer surface of the pulley are the same.
Now, to find the speed of block B after it has fallen 1.03 m, we need to find the acceleration of the system. We can do that with the sum of forces, and the torque about the pulley
For block A, Tension A pulls to the right, and friction (Fn) pulls to the left. The forces equation is...
TA - mAg = mAa
TA = (2.14)(a) + (.331)(2.14)(9.8)
TA = 6.94 + 2.14a
For block B, its weight pulls down while Tension B pulls upward. The forces equation is...
mBg - TB = mBa
TB = (6.85)(9.8) - (6.85)(a)
TB = 67.13 - 6.85a
For the pulley, we know that torque () = Fr. The forces are from the tension differences. We also know that torque = I. The equation becomes
(TB - TA)r = I
I for a disk is .5mr2 and = a/r
(TB - TA)r = .5mpr2a/r
Simplify
TB - TA = .5mpa
Substitute TB and TA
67.13 - 6.85a - 6.94 - 2.14a = (.5)(.421)(a)
Simplify
60.19 - 8.99a = .2105a
a = 6.542 m/s2
Now that we have a, we can apply the formula
vf2 = vo2 + 2ad
vf2 = 0 + (2)(6.542)(1.03)
vf = 3.67 m/s