A block slides along a frictionless surface with 220 J of kinetic energy. It then goes up a frictionless ramp. When the kinetic energy drops

Question

A block slides along a frictionless surface with 220 J of kinetic energy. It then goes up a frictionless ramp. When the kinetic energy drops to 55 J, how much gravitational potential energy does the block have?

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Delwyn 6 months 2021-07-26T22:16:06+00:00 1 Answers 2 views 0

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    2021-07-26T22:18:01+00:00

    Hello!

    For the explanation of this energy conservation exercise, where we’ll use energy conservation law, let’s see what this principle proposes.

    How you should know, mechanical energy conserves in every point, that is to say mechanical energy is same in A point like B point. (Mechanical energy will be represented by “Me”)

    Once time we know that, let’s take the 220 Joules momentum like A point, and when 55 Joules momentum like B point.

    Then, let’s use the energy conservation principle:

    Me(A) = Me(B)

    • We know Mechanical energy in A point, so just lets replace according to our data:

    220 J = Me(B)

    • In B point, we know kinetic energy, but we dont know gravitational potential energy, so lets descompose Mechanical energy, into kinetic energy and gravitational potential energy:

    220 J = Ke + Gpe

    • We know kinetic energy value, so lets replace it:

    220 J = 55 J + Gpe

    • Finally, just clean Gpe and resolve it:

    Gpe = 220 J – 55 J = 165 J

    Gravitational potential energy is of One hundred sixty five Joules (165 J).

                                                                                      ║Sincerely, ChizuruChan║

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