6. A 3.6-kg block lies on top of a second wooden block with a mass of 5.8-kg. Bo

Question

6. A 3.6-kg block lies on top of a second wooden block with a mass of 5.8-kg. Both masses initially are motionless on a frictionless table. The coefficient of static friction between the blocks is 0.90, and the coefficient of kinetic friction is 0.75. A person pushes with a force P on the lower block with the largest amount of force so that the blocks do not slip, relative to each other.

(i) What is the acceleration of the blocks

(ii) What is the force P?

(iii) What is the magnitude and direction of the NET force acting on the lower (5.8-kg) block from the upper

(3.6-kg) block?

A 3.6-kg block lies on top of a second wooden block with a mass of 5.8-kg. Both masses initially are motionless on a frictionless table. The coefficient of static friction between the blocks is 0.90, and the coefficient of kinetic friction is 0.75. A person pushes with a force P on the lower block with the largest amount of force so that the blocks do not slip, relative to each other. What is the acceleration of the blocks What is the force P? What is the magnitude and direction of the NET force acting on the lower (5.8-kg) block from the upper (3.6-kg) block?

Explanation / Answer

a) to not slip, top block has mass friction force

u mg = ma

a = ug = 0.9*9.81=8.829 m/s^2

b) for both blocks

P = (m1 + m2)a = (3.6+5.8)*8.829=82.99 N

c) Fy =- m g

Fx = umg

F = sqrt( (3.6*9.81)^2 + (0.9*3.6*9.81)^2)=47.51 N

direction = arctan(Fy/Fx) = arctan(1/u) = arctan(1/0.9)=-48.01 degrees

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