Asked by Bex
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.
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.
Answers
Answered by
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
Shouldn't the denominator be m1+m2? both are accelerating.
Net force=total mass*a
Answered by
nicholas
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
:)
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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