Physics

A 250g wood block firmly attached to a horizontal spring slides along a table with a coefficient of friction of 0.40. A force of 30.N compresses the spring 20cm. If the spring is released from this position what will be the block's speed when it passes equilibrium position?

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asked by Joey
  1. k = 30 N/.2 m = 150 N/m

    PE of spring = (1/2) k x^2 = 75(.04)=3 Joules

    normal force = m g
    friction force = .4 m g
    work done against friction = .2 *.4 * mg
    = .08*.25*9.81 = .196 Joules

    so Ke left = 3 - .196 = 2.8 Joules
    (1/2) m v^2 = 2.8
    calculate v

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    posted by Damon

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