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Potential Energy- Spring (10 points)?

A block is dropped onto a spring with k = 28 N/m. The block has a speed of 2.9 m/s just before it strikes the spring. If the spring compresses an amount 0.13 m before bringing the block to rest, what is the mass of the block?

The answer is 0.0432 kg, I need help getting there, please explain. Thank you.

2 Answers

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  • 8 years ago
    Favorite Answer

    The final potential energy of the spring is (1/2)kx^2 with x = 0.13 m.

    The energy that the block had which has been converted into that energy is (1/2)mv^2 + mgx. It had speed v = 2.9 m/sec when it touched the spring, but it falls an additional x = 0.13 m, gaining some more gravitational potential energy.

    Set (1/2)mv^2 + mgx = (1/2)kx^2 and plug in the known values of x, g, and k. Solve for m.

    That extra mgx term is something physics students often forget to account for when the spring is vertical.

  • ganz
    Lv 4
    4 years ago

    PE => skill potential PE = m * g * h, the position m => mass ( in kg) g => 9.8 m/s^2 h => height m and g being consistent, PE varies on the instant with h. So more suitable the h, more suitable the aptitude potential. PE is on the instant proportional to h. The golf ball has both energies. It has in effortless words Kinetic potential in effortless words previously hitting the floor. The bouncing causes lack of potential by using warmth era, and inelastic nature of the collision between the ball and the floor. The ball transformations direction, and loses Kinetic potential, and efficient aspects skill potential because it efficient aspects height.

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