A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 33 °. A 1.2 kg block is projected up the plane, fro

Question

A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 33 °. A 1.2 kg block is projected up the plane, from an initial position that is distance d = 0.90 m from the end of the relaxed spring, with an initial kinetic energy of 29 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.30 m? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.40 m?

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Phúc Điền 2 weeks 2021-07-19T13:46:51+00:00 1 Answers 2 views 0

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    2021-07-19T13:48:23+00:00

    Answer:

    Explanation:

    Initial kinetic energy = 29 J

    work done against gravity = mgsin33 x d  , m is mass of the block

    = 1.2 x 9.8 sin 33 x .9

    = 5.76 J

    potential energy stored in compressed spring

    = 1/2 k x², k is spring constant and x is compression

    = .5 x 200 x .3²

    = 9

    energy left = 29 – ( 5.76 + 9 )

    = 14.24 J

    b )

    energy stored in spring when compression is .4 m

    = 1/2 x 200 x .4²

    = 16 J

    required kinetic energy = 16 + 5.76

    = 21.76 J

    Block must be projected with energy of 21.76 J .

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