Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coeffieent of kinetic friction between block 2 and the table?
Answer: Let the tensions of on the strings connecting m2 and m3 be T23, and that connectiong m2 and m1 be T12, respectively. Applying Newton's second law with FN= m2g in this case to the system we have
m3g - T = m3a
T23 - µk m2g - T12 = m2a
T12 - m1g = m1a
Adding up the three equations and using m1 = M, m2 = m3 = 2M, we obtain
2Mg- 2µk Mg - Mg = 5 Ma
With a = 0.5m/s^2 this yields µk = 0.372. Thus, the coefficient of kinetic friction is roughly µk = 0.37
If you can, please explain and show haw they solved for µk because I didn't understand that last part. Thank you
2006-10-17
10:38:03
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1 answers
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asked by
gods1princesschanel
1