1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the work done in this case?

  1. Mar 19, 2017 #1
    1. The problem statement, all variables and given/known data
    A spring (k = 500 N/m) supports a 400 g mass in a vertical position. The spring is now stretched 15 cm and maintained at the stretched position. How much work is done in stretching the spring by 15 cm? Is this work dependent on the magnitude of force applied while stretching the spring?

    I am not sure if the work done would be just 0.5kx2 = 0.5 x 500 x 0.15[SUP}2[/SUP] = 5.625 J or one should use a more rigorous approach outlined under 3. The attempt at a solution.

    2. Relevant equations
    For a spring,
    F= kx
    PEspring = 0.5kx2

    Work energy Theorem
    ΔKE + Σ ΔPE = Work done

    3. The attempt at a solution
    Initial stretch of spring by 400 g mass = (0.400 x 9.8) / 500 = 0.00784 m
    The initial stretch is used to determine initial spring PE.

    Since the initial and final positions are at rest, so ΔKE = 0.
    Σ ΔPE = ΔPEspring + ΔPEgravity = (0.5 x 500 x (0.15 + 0.00784)2 - 0.5 x 500 x (0.00784)2) + (0 - 0.400 x 9.8 x 0.15 )
    = (6.23 - 0.015) + (- 0.588)
    = 5.627 J
    ∴ Work done = ΔKE + Σ ΔPE = 0 + 5.627 = 5.627 J

    This work would be independent of the magnitude of force applied since work done depends on only ΔKE and Σ ΔPE, both of which are independent of force applied.
  2. jcsd
  3. Mar 19, 2017 #2
    With only one significant figure in your original measurements, I don't think the difference in your answers will matter.
  4. Mar 19, 2017 #3
    So, the work in this context is force applied dotted with the displacement. You already recognize the force of the spring is ##F=kx##. We know that the displacement is 15 cm. Since the force changes as we stretch it further and further, we need to take the integral of the force with respect to x. When you do this, you'll end up with ##W=\frac 1 2 kx^2##.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted