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1 June, 01:01

A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a speed of 150 m/s and becomes embedded in the block. (a) Find the amplitude of the resulting SHM. (b) What percentage of the origi - nal kinetic energy of the bullet is transferred to mechanical energy of the oscillator

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  1. 1 June, 02:52
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    a. A = 0.1656 m

    b. % E = 1.219

    Explanation:

    Given

    mB = 4.0 kg, mb = 50.0 g = 0.05 kg, u₁ = 150 m/s, k = 500 N / m

    a.

    To find the amplitude of the resulting SHM using conserver energy

    ΔKe + ΔUg + ΔUs = 0

    ¹/₂ * m * v² - ¹/₂ * k * A² = 0

    A = √ mB * vₓ² / k

    vₓ = mb * u₁ / mb + mB

    vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

    A = √ 4.0 kg * (1.852 m/s) ² / (500 N / m)

    A = 0.1656 m

    b.

    The percentage of kinetic energy

    %E = Es / Ek

    Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

    Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

    % E = 13.72 / 1125 = 0.01219 * 100

    % E = 1.219
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