Physics work energy theorem help?

In a physics lab experiment, one end of a horizontal spring that obeys Hooke’'s law is attached to a wall. The spring is compressed 0.400 m, and a block with mass 0.300 kg is attached to it. The spring is then released, and the block moves along a horizontal surface. Electronic sensors measure the speed v of the block after it has traveled a distance d from its initial position against the compressed spring. The measured values are listed in (Figure 1) .

B)Use the work-energy theorem to derive an expression for v2 in terms of d.

Express your answer in terms of some or all of the variables k, m, d, μk and the acceleration due to gravity g.

1 Answer

  • The initial energy of the system is all in the spring:

    Eo = Uo = ½kx² = ½k(0.400)²

    for E and U in Joules and k in N/m.


    E = U + KE = Eo - work

    ½k(0.400 - d)² + ½mv² = ½k(0.400)² - µmgd → multiply by 2

    for d in meters, m in kg and v in m/s

    k(0.160 - 0.800d + d²) + mv² = 0.160k - 2µmgd

    kd(d - 0.800) + mv² = -2µmgd

    mv² = kd(0.800 - d) - 2µmgd

    v² = kd(0.800 - d)/m - 2µgd

    so if by "v2" you meant "v²," that's your answer.

    If you want the velocity, take the square root.

    Hope this helps!

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