A block of mass m1 = 4.50 kg sits on top of a second block of mass m2 = 15.4 kg,

Question

A block of mass m1 = 4.50 kg sits on top of a second block of mass m2 = 15.4 kg, which in turn is on a horizontal table. The coefficients of friction between the two blocks are s = 0.300 and k 0.100. The coefficients of friction between the lower block and the rough table are s = 0.500 and k 0.400. You apply a constant horizontal force to the lower block, just large enough to make this block start sliding out from between the upper block and the table (a) Draw a free-body diagram of each block, naming the forces on each. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Determine the magnitude of each force on each block at the instant when you have started pushing but motion has not yet started. In particular, what force must you apply? 110.74 |N (c) Determine the acceleration you measure for each block 98 1.96 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (m2) m/s- (m1

Explanation / Answer

c) Now, the acceleration for each block
block 1: a=m1*g*mu<k>/m1 = g*mu<k> = 9.8*0.1 = 0.98 m/s^2
for block2: net force actin on lower block is pushing force- force on block1, or
(m1+m2)*g*mus - m1*g*muk = (4.5 + 15.4)*9.8*0.5 - 4.5*9.8*0.4 = 79.87 N
and acceleration of the lower block = 79.87/15.4 = 5.186 m/s^2

Hope it is correct!!!

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