Question

Two masses, mA = 34.0 kg and mB = 40.0 kg , are connected by a rope that hangs over a pulley (as in the figure (Figure 1)). The center of the pulley is hollowed out so that you may assume all the mass of the pulley is in the rim. The radius of the pulley is 0.381 m and the mass of the pulley is 3.10 kg . Initially mA is on the ground and mB rests 2.50 m above the ground. g = 9.81 m/s2 .

If the system is released, use conservation of energy to determine the speed of mB just before it strikes the ground. Assume the pulley bearing is frictionless.

Review Two masses, mA = 34.0 kg and mB = 40.0 kg, are connected by a rope that hangs over a pulley (as in the figure (Figure 1)). The center of the pulley is hollowed out so that you may assume all the mass of the pulley is in the rim. The radius of the pulley is 0.381 m and the mass of the pulley is 3.10 kg. Initially mA is on the ground and mB rests 2.50 m above the ground. g = 9.81 m/s2 Part A If the system is released, use conservation of energy to determine the speed of mg just before it strikes the ground. Assume the pulley bearing is frictionless. П- Πνα ΑΣφ m/s Request Answer Submit Next Provide Feedback Figure 1 of 1 Ro тв mA P Pearson

Answer 1

Here we use energy conservation principle to find the required speed of the given mass as shown below***********************************************************************************************
Check the answer and let me know immediately if you find something wrong or missing... I will rectify the mistakes asap if any

- A As MB hits ground, work done by gravity is given by - Wgravity = -magh t mogh , put h=2.5m = 147:15J from conservation of energy DE = 0 => OU TAKE = 0 > - Wgravity FOKE = 0 fou = - Wgravity =) Wgravity = AKE = / MAV2 + 2 mg v² + 2 / I wa = { mave te movi+ { MRP) ? 141. 35 = // MAV2 + 2 / 2 mg v² + 2 /mv2 - put M = 3.11g of v = 1.95 m/s