Physics

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?

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  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

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  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

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