{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Final Exam Solutions

# From eq2 f 6250 9000 02512000 150045

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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,...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online