Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 years ago

Hooke s Law and Spring Potential Energy?

m of block = 1 Kg

K = 100 N/m

Box is stressed 25 inches (0.635 m) and then released.

Find:

1) The position of the block when the kinetic energy is equal to 1/4 the mechanical energy.

2) The velocity of the block when the potential energy is equal to zero.

3) The force on the block when the acceleration is equal to 1/2 the maximum acceleration.

It is a problem we were working on in class and I do not really understand it. Can someone please explain this?

1 Answer

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  • 3 years ago
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    Total Energy = 1/2*k*A^2 = 1/2*100 N/m *(.635 m)^2 = 20.2 J

    Energy at any point = total energy = kinetic energy + spring potential energy

    KE = 1/2*m*v^2

    SPE = 1/2*k*x^2

    1) KE = 1/4*Total Energy

    TE = 1/4*TE + SPE

    SPE = 3/4*TE

    1/2*k*x^2 = 3/4*TE

    X = sqrt (3/2*TE/k)

    X = sqrt (3/2*20.2 J / 100 N/m) = 0.55 m

    2) SPE = 0

    TE = KE = 1/2*m*v^2

    v = sqrt (2*TE/m)

    v = sqrt (2*20.2 J / 1 kg) = 6.36 m/s

    3) F_max = k*A = m*a_max

    a_max = k/m*A

    a_max = 100 N/m / 1 kg * 0.635 m = 63.5 m/s^2

    a_new = 1/2*a_max = 31.75 m/s^2

    F_new = m*a_new

    F_new = 1 kg * 31.75 m/s^2 = 31.75 N

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