One last physics question?

A block weighing 2 kg is pushed a spring with negligible mass and a spring constant of k=350 N/m, compressing it 0.25 m. If the block is released, it moves along a horizontal surface (frictionless) and then up an incline (frictionless) with a slope of 25 def.

a) What is the speed of the block as it slides across the flat surface after the spring?

b) How far can the block travel up the ramp before heading back down?

Thanks all, much appreciated.

2 Answers

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  • NCS
    Lv 7
    7 months ago
    Favorite Answer

    a) convert spring PE to KE

    ½kx² = ½mv²

    ½ * 350N/m * (0.25m)² = ½ * 2kg * v²

    solves to

    v = 3.3 m/s

    b) convert SPE to GPE

    ½kx² = mgh = mgLsinΘ

    ½ * 350N/m * (0.25m)² = 2kg * 9.8m/s² * L * sin25º

    solves to

    L = 1.32 m

    OR you could have converted the KE to GPE.

    Hope this helps!

    • NCS
      Lv 7
      7 months agoReport

      Thanks for awarding!

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  • 7 months ago

    I assume the slope is 25˚. The maximum kinetic energy of the block is equal to the work done on the spring.

    Work = ½ * 350 * 0.25^2 = 10.9375 N * m

    KE = ½ * 2 * v^2

    v^2 = 10.9375

    v = √10.9375

    This is approximately 3.3 m/s. The force that causes the block to decelerate is the component of its weight that is parallel to the inclined plane.

    Force parallel = 2 * 9.8 * sin 25 = 19.6 * sin 25

    The work done by this force is equal to the block’s maximum kinetic energy. Let d be the distance the block travel up the ramp before heading back down.

    Work = 19.6 * sin 25 * d

    19.6 * sin 25 * d = 10.9375

    d = 10.9375 ÷ 19.6 * sin 25

    The distance is approximately 1.3 meters.

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