A crate is given an initial speed of 5.0 m/s up the 22.5 ∘ plane shown in (Figure 1). Assume μk = 0.12.?

Question

How much time elapses before it returns to its starting point? (in seconds)

Answer 1

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Let's say the crate has a mass m

Upslope is the positive direction 

g = -9.81 m/s

θ = 22.5° from horizontal

Going up-slope

F = ma

mgsinθ + μmgcosθ = ma

a = g(sinθ + μcosθ)

a = -9.81(sin22.5 + 0.12cos22.5)

a = -4.8417 m/s²

v² = u² + 2as

s = (v² - u²) / 2a 

Going upslope, v = 0 and u = 5.0 m/s

s = - 5.0² / 2(-4.8417 m/s²)

The mass travels up-slope a distance of

s = 2.5817 m

s = ½at²

t₁ = √(2s/a)

t₁ = √(2(2.5817) / 4.8417

t₁ = 1.033  s

Using the same logic going downslope

a = -9.81(sin22.5 - 0.12cos22.5) 

a = - 0.2718 m/s²

t₂ = √(2(2.5817) / 0.2718 

t₂ = 4.358 s

total time is t₁ + t₂

t = 1.033 + 4.358

t = 5.391... ≈ 5.4 s

Answer 2

shown in (Figure 1) ???