Final Exam Solutions

0 kg the string does not slip on the rim of the

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Unformatted text preview: (KE + PE)A = (KE + PE)B M ! 10.0 ! 3.2 = 1 MVB2 or VB = 2 ! 10.0 ! 3.2 = 8.0 m / s 2 [b] (8 pts)What is the speed of M just after it exits the rough patch at C? (KE + PE)C = (KE + PE)final or KEC = PE of the compressed spring 1 1 MVc2 = kx 2 2 2 Vc = x k 2700 = 0.3 = 4.5m / s M 12.0 [c](6 pts) What is the co ­efficient of kinetic friction between the block and the rough patch BC? From B to C: ΔKEBC = Wnet = W f 1 (12.00)[(4.5) 2 ! (8.0) 2 ] = !262.5 = ! µ Mgd 2 µ= 262.5 262.5 = = 0.35 Mgd 750.0 N Ff Mg Prob.4 (20pts). An Atwood’s machine consists of mass m1 = 1.5 kg connected to mass m2 = 2.5 kg by a light, inextensible string that passes over a disc ­shaped pulley of radius R = 0.2 m and mass M =2.0 kg. The string does not slip on the rim of the pulley. The pulley rotates, without friction, about a horizontal axis passing through the center of the pulley. For parts [a] through [d] below, consider the instant at which m2 has fallen through a distance of 2.0 m, assuming it started from rest. P [a]( 8 pts) Using conservation of energy, find the speed of m2. (PE+KE)1= (PE+KE)2 Taking the gravitational potential energy to be zero at the starting position: 1 1 0 = m1 g (...
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This note was uploaded on 07/13/2013 for the course PHYS 101 taught by Professor Goldberg during the Summer '08 term at Drexel.

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