1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Homework Help: Spring Force and mass

  1. Apr 29, 2016 #1
    1. The problem statement, all variables and given/known data
    A 5.3kg mass hangs vertically from a spring with spring constant 720N/m. The mass is lifted upward and released. Calculate the force and acceleration the mass when the spring is compressed by 0.36m.
    Note: I already solved for acceleration and I got the correct answer- a=58.70566038m/s^2

    I tried to solve for the spring force but I got the wrong answer and I'm not sure what I did wrong.

    m=5.3kg, k=720N/m, Δx=0.36m.

    2. Relevant equations

    I used Fnet=ma
    3. The attempt at a solution

    The correct answer is 310N[down]. Can someone tell me what I did wrong? Also, I'm confused about whether a put the negative signs infront of the correct variables(Fg and Fx are both pointing downwards in this problem, so I put a negative sign infront of both). Should I put a negative sign infront of Fnet too since Fnet points downwards in this case? Are you even supposed to put negative signs infront of variables based on direction when doing calculations?
  2. jcsd
  3. Apr 29, 2016 #2
    It looks like you have some serious fundamental misunderstandings.

    Firstly, how'd you manage to find the acceleration of the mass without knowing the net force acting on the mass? What does Newton's Second Law tell us about how the net force acting on object is related to that object's acceleration?

    To answer your other question, yes you should keep track of your signs, you'll end up with wrong answer if you don't. In fact, your value of Fx is wrong because you're not using the right signs. Subtracting versus adding a quantity makes a big difference, so yes they are important.

    While we're at it, what force does Fx represent and why are you solving for it? The problem asks for the mass's acceleration, not the force by the spring.
  4. Apr 29, 2016 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you sure it is asking for the spring force, not the net force on the mass?
    If it does want the spring force, you have a sign wrong.
  5. Apr 29, 2016 #4
    i wonder how you got acceleration of the mass ;

    as usually when one draws a free body diagram the net force gets calculated and finally acceleration is equated with F/mass.
    mark the underlined question .... when spring is compressed...

    so the spring force will oppose compression and gravitational pull will be downward(opposed to compression), but the motion is upward; so calculate the net force which will be addition of the above two but acting opposed to displacement (effective deceleration)
    and if you divide the net force by mass you get acceleration.
    a look at the numbers gives an impression that you will get correct answer.
  6. Apr 30, 2016 #5
    It wants the spring force
  7. Apr 30, 2016 #6

    I did:


    I knew all the values except acceleration, so i just used algebra to solve for it
  8. Apr 30, 2016 #7
    It asks for 2 things: acceleration and spring force(Fx)

    Here's how I found acceleration:


    I knew all the above values except a, so I just used algebra to solve.

    Also, what mistake did I make with my negative signs?
  9. Apr 30, 2016 #8
    therefore you also knew the force as mass was known.
    by newtons laws if you know acceleration and mass then you know the force.
    so what value of force you get ?
  10. Apr 30, 2016 #9
    Fnet is 311.14N, which I should in my calculations in the original post
  11. Apr 30, 2016 #10
    so you are getting an approx. correct answer. regarding the negative sign - in force equations as they are vectors ,one has to use sign convention.
    suppose the body is going up and you are writing mass xacceleration = the net force '
    then the forces in the direction of motion may be taken as positive and opposed to the motion as negative.
    the sign of the accelaration after calculation can tell you whether the net force is helping the motion or otherwisei.e. retarding the motion.
    so the question you posed is perhaps asking for the net force rather than spring force;
    the wording of the question suggested that they need net force-
  12. Apr 30, 2016 #11
    What you posted says to find the force and acceleration of the mass. The two answers you provided agree with the values for both the force and the acceleration of the mass (F = -210N and a = -58). It also makes more sense that they'd ask for the force and then the acceleration of the mass (in that order) as what you posted suggests. Unless you're leaving something out, why do you think you need to find the spring force? It's impossible to help you if you don't post the entire problem exactly the way it was worded.

    Do note that -211.14 N is the same thing as 210 N (downward) once you take into account that the problem uses only two significant figures, so your final answer should only contain two significant figures.

    That aside, if you wanted to find the spring force, there's no need to work backward from the net force on the object. The spring force is given by Hooke's law and is independent of the acceleration of the mass and gravity. It only depends on how much the spring is compressed, which you know.

  13. Apr 30, 2016 #12
    It just asks for force and since the lesson was about the spring force, I assumed that that's what they wanted. Also, the only 2 forces acting on the mass are spring force and gravity, so it makes sense that they'd ask for the spring force.
  14. Apr 30, 2016 #13
    The question I posted is exactly as written in the textbook
  15. Apr 30, 2016 #14
    So, I should put a negative sign infront of the acceleration too
  16. Apr 30, 2016 #15
    I didn't get F=210N, I got 363.08N
  17. Apr 30, 2016 #16
    you do not have to put in a sign- its the result of your calculation that acceleration = - number m/s^2
    and this carries a meaning that it is in a direction opposite to the motion.
  18. Apr 30, 2016 #17
    Oh ok, so then I put the negative signs in the correct places, but still got the wrong answer
  19. Apr 30, 2016 #18
    no you got 311 N for the net force and its close to 310 which was expected.
    It is the spring force pushing downward because of compression of spring , there was no term of 210 N in the discussion.and 363 is the wrong answer.

  20. Apr 30, 2016 #19
    Why does spring force=net force? or does the question just want us to solve for net force?
  21. Apr 30, 2016 #20
    The problem asks you to find the net force acting on the mass and the mass's acceleration. The answers given are for the force and acceleration of the mass.

    I don't understand why you think you're finding the spring force. Even then, you can't find the correct value for it because you aren't using the acceleration you claimed to have found, which is a claim I'm now very skeptical of. In fact, you had to use the spring force in order to find the acceleration, so the fact that you're having trouble finding it is concerning.
    Last edited: Apr 30, 2016
  22. Apr 30, 2016 #21
    i do not know whether you have done simple hamonic motion with a mass hanging on the spring;
    i quote a simple way to look at it

    the spring is normally supporting a mass by some stretching increase in length - any disturbance leads to oscillations ,if it is during compression stage the net downward force will be provided by the spring as all the time it is holding the mass
  23. Apr 30, 2016 #22
    Got ahead of myself.
  24. Apr 30, 2016 #23
    you are right about the force mg acting downward but in oscillatory motion the equilibrium point gets shifted and the initial condition supports the mass
    Assume a mass suspended from a vertical spring of spring constant k. In equilibrium the spring is stretched a distance x0 = mg/k. If the mass is displaced from equilibrium position downward and the spring is stretched an additional distance x, then the total force on the mass is mg - k(x0 + x) = -kx directed towards the equilibrium position. If the mass is displaced upward by a distance x, then the total force on the mass is mg - k(x0 - x) = kx, directed towards the equilibrium position. The mass will execute simple harmonic motion. The angular frequency ω = SQRT(k/m) is the same for the mass oscillating on the spring in a vertical or horizontal position. The equilibrium length of the spring about which it oscillates is different for the vertical position and the horizontal position.
  25. May 1, 2016 #24
    Well, I did find the acceleration first and I got the correct answer, so I don't understand why that's hard to believe
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted