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Homework Help: Spring/Force Problem

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant k inside a box. The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box?

    Diagram: http://www.aapt.org/Programs/contests/upload/olympiad_2008_fnet_ma.pdf
    Problem 17
    (Answer: ma/k)

    2. Relevant equations
    F = ma
    F = -kx

    3. The attempt at a solution
    Well, the forces acting are: the force of the spring (Fs), the weight of the object(Fg), and the force of the elevator (Fe).
    Fe + fs - mg = ma
    Fe - kx = m (a+g)
    I thought this was correct but am not sure where to progress from here.
  2. jcsd
  3. Jan 27, 2013 #2


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    Be careful when specifying the forces that act on the mass m. Which of the three forces that you listed does not act on m?
  4. Jan 27, 2013 #3
    Is the equation simply:
    F_s = ma, where a is acceleration of box,
    -kx = ma
    x = ma/k
    Last edited: Jan 27, 2013
  5. Jan 27, 2013 #4
    @Signature PF:
    You might have made an error, because you forgot about the negative, it'd be x = - ma / k

    Fe i suppose does not act directly on m, but the weight DOES directly affect the mass m.
    The way the problem is worded, the weight overcame the force of the spring.
    mg - (-kx) = ma
    x = m (a-g) / k,
    so I am still making an error somewhere
  6. Jan 27, 2013 #5
    Morris - the sign is insignificant in this case, all it represents is compression vs extension. In this case, the negative is understood to mean that the spring is compressed, so I omitted it in the answer.
  7. Jan 27, 2013 #6


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    Right, the only two forces acting on m are the force of gravity and the spring force.

    You will need to be careful with the signs. Since the box is accelerating upward, it might be good to take upward as the positive direction. So gravity exerts a force of magnitude mg downward while the spring force exerts a force of magnitude kx upward. When you fix the signs and solve for x, you will want to compare the result to the value of x when the box is not accelerating.
  8. Jan 27, 2013 #7
    When the box is not accelerating:
    kx = mg
    x = mg / k

    When the box has an acceleration a:
    kx - mg = ma
    kx = mg + ma
    x = m(g+a) / k
    Subtracting the two yields:
    mg + ma - mg / k
    = ma / k
    This is the desired answer.
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