## A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline,

Question

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.60 m to the bottom of the incline is 3.80 m/s. What is the speed of the block when it is 4.60 m from the top of the incline

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6 months 2021-07-22T07:19:31+00:00 1 Answers 6 views 0

h ’= 0.51356 m

Explanation:

For this exercise we can use conservation of energy

starting point. Highest point of the trajectory

Em₀ = m g h

sin θ = h / L

h = L sin θ

Em₀ = m g L sin  θ

final point. Lowest point of the trajectory

Em_f = K = ½ mv²

as there is no friction, energy is conserved

Em₀ = Emf

m g L sin  θ  = ½ m v²

sin  θ  = ½ v² / gL

sin  θ  = ½ 3.80² / (9.8  6.60)

sin  θ  = 0.111626

tea = 6.41

They ask us for the speed for L ’= 4.60 m

let’s find the height

sin 6.41 = h ‘/ L’

h ’= L’ sin 6.41

h ’= 4.60 sin 6.41

h ’= 0.51356 m

we use conservation of energy

m g h ’= ½ m v’2

v ’= √ (2gh’)

v ’= √ (2 9.8 0.51356)

v ’= 3.173 m /s²