Posted by **Zac** on Wednesday, July 25, 2012 at 8:24pm.

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 0.300, and the surface on which the 8.00-

kg block rests is frictionless. A constant horizontal force of magnitude F = 10.0 N is applied to

the 2.00-kg block, setting it in motion across the top of the lower block. If the distance across the

larger block is 3.00 m (from front edge of smaller block to rightmost edge of larger block),

(a) how long will it take the smaller block make it to the right side of the 8.00-kg block. (b) How

far will the 8.00-kg block move in this time?

- Physics -
**Elena**, Thursday, July 26, 2012 at 3:37pm
m1=2 kg, m2=8 kg, μ=0.3, F=10 N.

For m1:

m1•g=N,

m1•a1=F-F(fr) = F- μ•N=F- μ•m1•g.

a1=F/m1 - μ•g = 10/2 -0.3•9.8 = 2.06 m/s²

For m2:

m2•a2=F(fr)

a2=F(fr)/m2= μ•m1•g/m2 = 0.3•2•9.8/8 = 0.735 m/s².

Distances

x1=a1•t²/2,

x2=a2•t²/2,

x1=x2+L,

a1•t²/2 = a2•t²/2 + L,

Solve for t

t=2.13 s.

x2=a2•t²/2= 1.67 m

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