Question

A mass (m1) is connected by a light string that passes over a massless frictionless pulley to a mass (m2) sliding on a horizontal surface. The coefficient of kinetic friction between the m2 and the surface is 0.25. If m1=1.0kg and m2=2.0kg then what is the acceleration of the system?
I know I need to use the equation:
a= [(m1g) - (μk)(m2g)]/(m1-m2).
The answer for my acceleration is supposed to equal 1.6

*I'm having trouble identifying/obtaining the value for μk.*
I know Fk=μk(Fn), and Fn=(m2)(g)=19.6

I tried taking the answer for acceleration and solving for the μk value I'm supposed to have and I got 0.41836735, but I have no idea how I would get that number to solve for the acceleration to begin with.
bobpursley
mk is given as .25, the kinetic force of friction constant.

Shouldn't the denominator be m1+m2? both are accelerating.

Net force=total mass*a
total mass = m1+m2 = 3

down force on m1 = 9.8 (m1.g)
fric force on m2 = 2*9.8/4 = 9.8/2
(m2.g.mk, cuz mk = 0.25 = 1/4)

net force = 9.8 - 9.8/2 = 9.8/2

a = net force/total mass = 9.8/2/3 = 1.6

:)

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