The system shown in the figure below consists of a m1 = 5.52-kg block resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging m2 = 2.76-kg block.The pulley is a uniform disk of radius 7.86 cm and mass 0.592 kg. Calculate the speed of the m2 = 2.76-kg block after it is released from rest and falls a distance of 2.08 m. Calculate the angular speed of the pulley at this instant.
s=2.08 m
m=0.592 kg
R=0.0786 m
a is the acceleration of the system (m1+m2)
m₁a= T₁
m₂a= m₂g-T₂
Iε=M => mR²•a/2R = (T₂-T₁)R,
T₂-T₁= ma/2,
a= m₂g/( m₁+m₂+m/2)=...
a=v²/2s
v=sqrt(2as)=...
ω=v/R=...
To solve this problem, we can use the principles of conservation of energy and the relationship between linear and angular motion. Here's how you can calculate the speed of the 2.76 kg block and the angular speed of the pulley:
1. Calculate the potential energy of the hanging block:
Potential energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
PE = 2.76 kg * 9.8 m/s^2 * 2.08 m
2. Calculate the change in potential energy as the block falls:
ΔPE = PE
3. Calculate the kinetic energy of the falling block:
Kinetic energy (KE) = 1/2 * mass (m) * velocity^2
KE = 1/2 * 2.76 kg * velocity^2
4. Set the change in potential energy equal to the kinetic energy:
ΔPE = KE
2.76 kg * 9.8 m/s^2 * 2.08 m = 1/2 * 2.76 kg * velocity^2
5. Solve for the velocity of the block:
velocity = √(2 * ΔPE / m)
velocity = √(2 * (2.76 kg * 9.8 m/s^2 * 2.08 m) / 2.76 kg)
6. Calculate the angular displacement of the pulley:
Angular displacement (θ) = (distance traveled by block) / (radius of pulley)
θ = 2.08 m / 0.0786 m (converted cm to m)
7. Calculate the angular speed of the pulley using the relationship between linear and angular motion:
Angular speed (ω) = velocity / radius of pulley
ω = velocity / 0.0786 m (converted cm to m)
After following these steps, you should obtain the speed of the 2.76 kg block and the angular speed of the pulley.