Question

If the crate starts from rest at height of 8.15 m from base of the plane, what will be the crate's speed when it reaches the bottom of the incline? lies on a plane tilted at an angle θ = 25 ∘ to the horizontal, with μk = 0.19

Answer 1

Answer:

9.73 m/s

Explanation:

Draw a free body diagram of the crate. There are 3 forces:

Weight force mg pulling down,

Normal force N pushing up perpendicular to the incline,

and friction force Nμ pushing up parallel to the incline.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of forces in the parallel direction:

∑F = ma

mg sin θ − Nμ = ma

mg sin θ − mgμ cos θ = ma

g sin θ − gμ cos θ = a

a = g (sin θ − μ cos θ)

Plug in values:

a = (9.8) (sin 25° − 0.19 cos 25°)

a = 2.45 m/s²

Given in the parallel direction:

s = 8.15 / sin 25° = 19.3 m

u = 0 m/s

a = 2.45 m/s²

Find: v

v² = u² + 2as

v² = (0)² + 2 (2.45) (19.3)

v = 9.73 m/s