A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 k

Question

A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 kg. The coefficient of kinetic friction between the two blocks is 0.345, and the surface on which the bottom block rests is frictionless. A constant horizontal force of magnitude F = 14.0 N is applied to the top block, setting it in motion as shown in figure (a). The distance that the leading edge of the smaller block travels on the larger block is L=3.00 m.

(a) What is the acceleration of the top block?

(b) What is the acceleration of the bottom block?

(c) In what time interval will the top block make it to the right side of the bottom block as shown in figure (b)?

(d) How far does the bottom block move in the process?

Explanation / Answer

t = (2x/(a1-a2))
A)acceleration of top block a1 = (F-Ff)/m 1 = (14 - 2gµ)/2 = 3.619 m/s²
B)acceleration of bottom block a2= Ff/m 2 = 2gµ/m2 = 6.762/8 = 0.845 m/s²

C) time interval, t = (2 x 3/(3.619 - 0.845)) = 1.470 sec

D) bottom block will move,

x2= 1/2(a2 x t 2) = 1/2(0.845 x 1.470² )= 0.912 m

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