A block is at rest on the incline shown in the figure. FREE 10 PTS if its correct!?

A)

friction force is

F=μsmgcosθ

F=0.67*36*9.8*cos30

F=24.7 N

B)

at rest on have

0=mgsinθ-μsmgcosθ

then

tanθ=μs

θ=arctan0.67=33.82 degrees

C)

ma=mgsinθ-μdmgcosθ

a=g(sinθ-μdcosθ)

a=9.8(sin37-0.57cos37)

a=1.44 m/s^2

The block has two forces which affect its acceleration. The component of block’s weight which is parallel to the incline is causing it to accelerate. The friction force is causing it to decelerate. Since the block is at rest. The force parallel is equal to the friction force. Since the block is at rest, use the coefficient of static friction.

Ff = 0.67 * 36 * 9.8 * cos 30 = 236.376 N

To determine the largest angle, set the force parallel equal to the friction force.

Fp = 36 * 9.8 * sin θ

Ff = 0.67 * 36 * 9.8 * cos θ

36 * 9.8 * sin θ = 0.67 * 36 * 9.8 * cos θ

Divide both sides by 36 * 9.8 * cos θ

Tan θ = 0.67

The angle is approximately 33.82˚

Use the following equation to determine the acceleration when the angle is 37˚

Net force = mass * acceleration

Net force = Force parallel – Friction force

Since the block is accelerating, use the coefficient of kinetic friction.

Fp = 36 * 9.8 * sin 37 = 352.8 * sin 37

Ff = 0.57 * 36 * 9.8 * cos 37 = 201.096 * cos 37

352.8 * sin 37 – 201.096 * cos 37 = 36 * a

a = (352.8 * sin 37 – 201.096 * cos 37) ÷ 36 = 9.8 * sin 37 – 5.586 * cos 37

The acceleration is approximately 1.4366 m/s^2