Find the acceleration experienced by each of the two objects , if the coefficient of kinetic friction between the 7kg object and the plane is 0.25

Explanation:

Find the acceleration reached by each of the two objects shown in Figure P4.49 if the coefficient of kinetic friction between the 7.00 kg object and the plane is 0.270.

The weight of the 12.00 kg object = 12.00 * 9.8 = 117.6 N

This force is accelerating the 7.00 kg object up the incline, and accelerating the 12.00 kg object straight down.

The weight of the 7.00 kg object = 7.00 * 9.8 = 68.6 N

This force has causes the 7.00 kg object to accelerate down the incline, and produces a force perpendicular to the surface of the incline.

The force perpendicular to the surface of the incline produces the Friction force that resists motion down the incline.

Force parallel = mass * g * sin θ = 7.00 * 9.8 * sin 37º = 68.6 * sin 37º

Force perpendicular = mass * g * cos θ  

Friction force = μ * mass * g * cos θ = 0.270 * 7.00 * 9.8 * cos 37º

Friction force = 18.522 * cos 37º

Force caused by weight of 12.00 kg object = 117.6 N

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

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