A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it with a hammer

Question

A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 36.0 cm/s .
What is the amplitude of the subsequent oscillations?
What is the block’s speed at the point where x= 0.70 A?

in progress 0
Edana Edana 4 years 2021-07-22T15:16:25+00:00 1 Answers 23 views 0

Answers ( )

  1. Amity
    0
    2021-07-22T15:17:25+00:00
    Reply

    Answer:

    Explanation:

    Given that,

    Mass attached m = 0.95kg

    Spring constant k = 16N/m

    Instantaneous speed v = 36cm/s = 0.36m/s

    Amplitude A=?

    When x = 0.7A

    Using conservation of energy

    ∆K.E + ∆P.E = 0

    K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

    At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

    Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

    So, the equation becomes

    — K.E(initial) + P.E(final) = 0

    K.E(initial) = P.E(final)

    ½mv² = ½kA²

    mv² = kA²

    0.95 × 0.36² = 16×A²

    0.12312 = 16•A²

    A² = 0.12312/16

    A² = 0.007695

    A = √0.007695

    A = 0.088 m

    A = 8.8cm

    B. Speed at x = 0.7A

    Using the same principle above

    K.E(initial) = P.E(final)

    ½mv² = ½kA²

    Where A = 0.7A = 0.7 × 0.088 = 0.0614m

    Then,

    ½× 0.95 × v² = ½ × 16 × 0.0614²

    0.475v² = 0.0310644

    v² = 0.0310644/0.475

    v² = 0.0635

    v = √0.0635

    v = 0.252 m/s

    v = 25.2 cm/s

Leave an answer

Browse

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