Question

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Answer 1

Answer:

63.8

Explanation:

Answer 2

Answer:

70 N

21°

1.1 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

Weight pulling straight down

Normal force pushing perpendicular to the incline

Friction force pushing parallel to the incline

Part 1

Sum the forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

The block is at rest, so F = N μs:

F = N μs

F = mg μs cos θ

F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)

F = 70 N

Part 2

Sum the forces in the parallel direction (down the incline is positive):

∑F = ma

mg sin θ − F = 0

mg sin θ = N μs

mg sin θ = mg μs cos θ

tan θ = μs

θ = atan μs

θ = atan 0.38

θ = 21°

Part 3

Sum the forces in the parallel direction (this time, acceleration is not 0).

∑F = ma

mg sin θ − F = ma

mg sin θ − N μk = ma

mg sin θ − mg μk cos θ = ma

a = g (sin θ − μk cos θ)

a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)

a = 1.1 m/s²