A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a distance 2.80

Question

A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a distance 2.80 × 10⁻² m. The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring.
Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. What is the speed of the block after it leaves the spring?

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bonexptip 3 years 2021-09-05T14:36:30+00:00 1 Answers 13 views 0

Answers ( )

  1. Vodka
    0
    2021-09-05T14:37:39+00:00
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    Answer:

    W = 0.060 J

    v_2 = 0.18 m/s

    Explanation:

    solution:

    for the spring:

    W = 1/2*k*x_1^2 – 1/2*k*x_2^2

    x_1 = -0.025 m and x_2 = 0

    W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

    W = 0.060 J

    the work-energy theorem,

    W_tot = K_2 – K_1 = ΔK

    with K = 1/2*m*v^2

    v_2 = √2*W/m

    v_2 = 0.18 m/s

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