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Homework Help: Need confirmation for spring mass system question

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass of 1.20 kg. originally at rest, sits on a frictionless surface. It is attached to one end of an unstretched spring (k = 790 N/m), the other end of which is fixed to a wall. The mass is then pulled with a constant force to stretch the spring. As a result, the system comes to a momentary stop after the mass moves 14.0 cm.

    Find the new equilibrium position of the system.

    2. Relevant equations
    Total Work = Change in kinetic energy = 0
    W = F * d
    Total Work = Work done by force + Work done spring

    3. The attempt at a solution

    My first thought was that since the mass comes to a stop in 14 cm then that must be the equilibrium position since it would only come to a stop if the spring force equaled the applied force.

    However the answer is actually found by first calculating the work done by the applied force:

    Total Work = Work by Force + Work by Spring = Change in KE = 0
    Work by Force = - Work by Spring
    Work by Force = 7.4 J

    Then you determine the magnitude of the applied force:

    W = Force * Displacement
    W / Displacement = Force
    Force = 55 N

    Then you determine when the Spring Force is equal to 55 N:

    55 = (0.5) * k * x^2
    55 = (0.5) * 790 * x^2
    x = 0.07 m = 7 cm

    Now, my confusion arises from the fact that in the question it is stated that the spring comes to a MOMENTARY STOP at 14 cm. How is that possible if the applied force is equal to the spring force at 7 cm? How is the mass able to travel an additional 7 cm before coming to a momentary stop?

    Is it due to the fact that when it reaches 7 cm it has an initial velocity and that kinetic energy is converted to spring energy over a distance of 7 cm? Okay I just checked the math and this works out perfectly. So I guess this answers my question but I would love if someone could confirm this. Thank you for reading.
  2. jcsd
  3. Oct 24, 2015 #2

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    How are you getting 7.4 J?

    ##\frac{1}{2}kx^2## is an energy, not a force. In SI units it gives you a number in joules, not newtons.
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