Consider the system shown in the figure below. Block A weighs 49.6 N and block B

Question

Consider the system shown in the figure below. Block A weighs 49.6 N and block B weighs 28.2 N. Once block B is set into downward motion, it descends at a constant speed.

magnitude m/s2 Consider the system shown in the figure below. Block A weighs 49.6 N and block B weighs 28.2 N. Once block B is set into downward motion, it descends at a constant speed. Calculate the coefficient of kinetic friction between block A and the tabletop. A cat, also of weight 49.6 N, falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration?

Explanation / Answer

a constant speed means no acceleration so ,

m1a1=m2a2

Mag -T = T-mN

Mbg=mMag

m=coefficient of friction kinetic = 28.2/49.6 = 0.568 m/sec^2

b . so , T-mN= MaAa

Mbg-T = MbAb , and Aa=Ab they are connected right

add two equations , we get 28.2 g - 0.568 Mag = Ma Aa

Acceleration = 2.7805 m/s^2

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