A stationary 2.0 kg object is located on a table near the surface of the earth. The coefficient static friction between the surface is .80 and the coefficient of the kinetic friction is 0.65. A horizontal force of 5N is applied to the object. Determine the force of friction. Determine the accelaration of the object. How should we change the magnitude of the horizontal applied force to get an acceleration of 1.5 m/s^2? What is the friction force at that acceleration?

Answer 1

m g = weight = 2*9.81 = 19.62 N

static friction max = .8*19.62 = 15.7 N
kinetic friction = .65*19.62 = 12.8 N

5 N does not move it. Friction resists with 5 N by third law

F - 12.8 = 2 (1.5)
F = 15.8 N to get 1.5 m/s^3 acceleration, it is moving now so friction force is -12.8 as soon as it starts

Answer 2

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