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fallen186
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**Sorry the latex didn't really work for me.**
In the figure below, m1 = 3.6 kg, m2 = 5.1 kg, and the coefficient of kinetic friction between the inclined plane and the 3.6-kg block is μk = 0.25. Find the magnitude of the acceleration of the masses.
m1 = 3.6 kg
m2 = 5.1 kg
μ = 0.25
Theta = 30
a = ((m2*g)-(m1*g*sin(theta)))/(m1+m2)
Since there is friction I did [tex] (m1*g*sin(theta)-μ*Fn
Fn = (.5*m1*g*3^.5)
a = ((5.1 kg*9.8 m/s^2) - (3.6 kg*9.8 m/s^2*sin(30)-.25(.5*3.6kg*9.8m/s^2*3^.5)))/(3.6kg+5.1kg)
a = (50.031- (17.658 - 7.638))/(8.7)
a = 4.599 m/s^2
I got this wrong. please help
Homework Statement
In the figure below, m1 = 3.6 kg, m2 = 5.1 kg, and the coefficient of kinetic friction between the inclined plane and the 3.6-kg block is μk = 0.25. Find the magnitude of the acceleration of the masses.
m1 = 3.6 kg
m2 = 5.1 kg
μ = 0.25
Theta = 30
Homework Equations
a = ((m2*g)-(m1*g*sin(theta)))/(m1+m2)
The Attempt at a Solution
Since there is friction I did [tex] (m1*g*sin(theta)-μ*Fn
Fn = (.5*m1*g*3^.5)
a = ((5.1 kg*9.8 m/s^2) - (3.6 kg*9.8 m/s^2*sin(30)-.25(.5*3.6kg*9.8m/s^2*3^.5)))/(3.6kg+5.1kg)
a = (50.031- (17.658 - 7.638))/(8.7)
a = 4.599 m/s^2
I got this wrong. please help