Final Exam Solutions

From eq2 f 6250 9000 02512000 150045

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Unformatted text preview: es. Take the x ­axis along the incline. For m: N1= 500.0cos37o = 400.0N T – 500.0sin37 – Ff = 50.0a …[1] For M: N2 = 1500.0 cos37 = 1200.0 N F – T – 1500sin37 – Ff’ = 150.0a …[2] [c] (4 pts). What is the maximum acceleration with which m and M can be pulled up the incline? From eq.[1] with T = 625.0N and Ff = µN1 625.0 – 300.0 – 0.25*(400.0) = 50.0 a or a = 225.0/50.0 = 4.5 m/s2 [d] (2 pts).What is the maximum force F that can be applied without breaking the connecting rope? From eq.[2] F = 625.0 + 900.0 + 0.25(1200.0) + 150.0*4.5 = 2500.0 N Prob. 3(20 pts). A block of mass M = 12.0kg is released from rest at point A as shown below(h = 3.2m). The track is frictionless except for the portion BC of length d = 6.25m. The block moves down the track, hits a spring of spring constant k = 2700 N/m, and compresses the spring 0.3m from its equilibrium position before coming to rest momentarily. M A g h d C B [a](6 pts) What is the speed of M just before it enters the rough patch at B? using,...
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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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