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 0.300, 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?

Answer 1

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

Answer 2

given, m1 = 2kg ; m2 = 8kg; coefficient of friction = 0.3; F = 10 N

For m1:
m1*g = N
m1*a1 = F-F(friction) = F- (coefficient of friction)*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(friction)
a2 = F(friction)/m2= μ*m1*g/m2 = 0.3*2*9.8/8 = 0.735 m/s²
Distance
x1 = a1*t^2/2
x2 = a2*t^2/2
x1 = x2+L,
a1*t^2/2 = a2*t^2/2 + L
Solving for t,
t = 2.13 s
x2 = a2*t^2/2 = 1.67 m
Answer