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pcandrepair
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[SOLVED] Speed of a Block
The sliding block has a mass of 0.830 kg, the counterweight has a mass of 0.430 kg, and the pulley is a hollow cylinder with a mass of 0.350 kg, an inner radius of 0.020 m, and an outer radius of 0.030 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward the pulley when it passes through a photogate.
(a) Use energy methods to predict its speed after it has moved to a second photogate, 0.700 m away.
_________ m/s
(b) Find the angular speed of the pulley at the same moment.
_________ rad/s
Moment of Inertia for a hollow sphere: Icm = 1/2 M (R1^2 + R2^2)
Friction = [tex]\mu[/tex]*M1*g*diameter
For part A i used the following equation to solve for the final velocity:
1/2(M1*Vi^2) + 1/2(M2*Vi^2) - friction = 1/2(M1*Vf^2) + 1/2(M2*Vf^2) + M2*g*h
1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m)
-1 = 1/2 Vf^2 (.83Kg + .43Kg) + 2.9498
Vf^2 = -4.5371
Vf = 2.13 m/s
This answer is incorrect. I cannot figure out where I went wrong. If someone could help me out, it would be much appreciated. Thanks!
I also tried part b with this number. I'm pretty sure this is the correct formula but I need the right number from part a to get the correct answer.
V = R[tex]\omega[/tex]
[tex]\omega[/tex] = V/R
[tex]\omega[/tex] = (2.13m/s) / (sqrt((.02^2) + (.03^2)))
[tex]\omega[/tex] = 59.0756 radians/sec
Homework Statement
The sliding block has a mass of 0.830 kg, the counterweight has a mass of 0.430 kg, and the pulley is a hollow cylinder with a mass of 0.350 kg, an inner radius of 0.020 m, and an outer radius of 0.030 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward the pulley when it passes through a photogate.
(a) Use energy methods to predict its speed after it has moved to a second photogate, 0.700 m away.
_________ m/s
(b) Find the angular speed of the pulley at the same moment.
_________ rad/s
Homework Equations
Moment of Inertia for a hollow sphere: Icm = 1/2 M (R1^2 + R2^2)
Friction = [tex]\mu[/tex]*M1*g*diameter
The Attempt at a Solution
For part A i used the following equation to solve for the final velocity:
1/2(M1*Vi^2) + 1/2(M2*Vi^2) - friction = 1/2(M1*Vf^2) + 1/2(M2*Vf^2) + M2*g*h
1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m)
-1 = 1/2 Vf^2 (.83Kg + .43Kg) + 2.9498
Vf^2 = -4.5371
Vf = 2.13 m/s
This answer is incorrect. I cannot figure out where I went wrong. If someone could help me out, it would be much appreciated. Thanks!
I also tried part b with this number. I'm pretty sure this is the correct formula but I need the right number from part a to get the correct answer.
V = R[tex]\omega[/tex]
[tex]\omega[/tex] = V/R
[tex]\omega[/tex] = (2.13m/s) / (sqrt((.02^2) + (.03^2)))
[tex]\omega[/tex] = 59.0756 radians/sec