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

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

Answers ( )

    0
    2021-07-22T07:20:33+00:00

    Answer:

       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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )