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!

Spring constant

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A block of mass 1 kg is attached to a spring. The spring extends by 10 cm. Find spring constant.


    2. Relevant equations


    3. The attempt at a solution
    Potential energy of spring = kx2/2
    work done by block = PE
    Hence
    mg*x=kx2/2
    ∴1*9.8*0.1=k*0.1*0.1/2
    ∴k=196N/m
    But solution says
    mg = kx (in magnitude )
    ∴k=mg/x=98N/m
     
    Last edited by a moderator: Mar 1, 2016
  2. jcsd
  3. Mar 1, 2016 #2
    Is this a vertical spring? What are the forces acting on it?
     
  4. Mar 1, 2016 #3
    Is there any information about the 'spring'- if the spring is 'massless' then one can expect the energy of extension by external mass/force to be same-otherwise when one hangs a spring having mass ,it will extend by its own weight.(in a very slow process) a dynamical process.

    another point is that if the body has converted a P.E. of mgx by falling through height x then a part of it must go in K.E. of the system and it must have fallen to larger x-value and oscillated before reaching an equilibrium and ultimately an equilibrium has reached with mean extension.
    so thermal energy loss must be there.
    try to think over it!

    so the extension produced by the wt mg can be seen as static equilibrium of the body - a free body diagram equates mg -equal to the restoring force proportional to extension.
     
  5. Mar 1, 2016 #4
    Altho it doesn't specify it in the description, the assumption is that said spring is hanging vertically,
    so the weight of the block = mg is the force that extends the spring a distance = 10 cm.
    Hooke's Law: Spring's force (mg) is directly proportional to the extension distance: F = kx
    {where F = force, k = spring constant, and x = spring length extension}
    F = mg = kx
    k = mg/x = 1(9.8)/0.1 = 98 N/m ANS
    Altho an energy approach could be used to solve, it is NOT a two term equation of simply SPE = GPE.
    The block moves vertically down a distance = x = 0.10 m, BUT it doesn't Free-Fall that distance.
    It falls with the force of the spring pulling UP against it's weight. So negative work is done by the spring force on the block.
    GPE + Spring Work = 1/2kx²
    mgh - Favg • h = 1/2kx² {where h = x = 0.10 m}
    Favg = (9.8 + 0)/2 = 4.9 N
    9.8(0.10) - 4.9(0.10) = 1/2k(0.10)²
    0.49 = (0.5)k(0.01)
    k = 98 N/m
     
  6. Mar 1, 2016 #5
    I am quite interested to know how you assumed this. I could as well generate a gravitational field in a horizontal field and compress the spring. Or the spring could be vertical and in a tub of liquid helium. Or else the whole thing could be floating in artificial vacuum generated in the Empire State Building. Let the OP verify my friend.

    @Prathamesh ... could you specify if any other info is given? Is the spring vertical? Horizontal with an external force? What are the conditions?
     
  7. Mar 2, 2016 #6
    Spring is massless and vertical
     
  8. Mar 2, 2016 #7
    I dont understand the concept of spring work...
    Is this concept introduced due to variable force with respect to x?
     
  9. Mar 2, 2016 #8
    2

    If you calculate the work done by the force during its elongation by taking an element at say y and y+dy then the work done = f(net). dy
    F(net) can be taken as mg acting down ward and -Ky acting upward
    so F(net)= mg- Ky Work done = integral of (mg-Ky).dy between limits L and L+x or effectively zero to elongation x.
    then the energy stored = mgx- (1/2) Kx^2 ; perhaps one can say that negative work is being done by the spring force. pl. think over it.
     
  10. Mar 2, 2016 #9
    I've one more problem. But if it is solved , by 1st method (F=kx) then ans comes wrong..... SPRING.jpg

    Hence,
    Tension in string = mgsin37-μmgcos37 which extends/pulls spring of k = 100N/m by 0.1m
    F=kx=100*0.1=10N
    ∴mgsin37-μmgcos37=10
    But in this way μ comes negative.
    And solution says
    Potential ene. of spring = work done by the block
    ∴1/2kx2=(mgsin37-μmgcos37)x
    and then ans comes....
    Then where is spring work????
    Plzzz help me.
     

    Attached Files:

  11. Mar 2, 2016 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Your difficulty is because the question is unclear.
    Merely attaching a mass to a spring, even if the spring is vertical, does not extend it. Something else has to happen, and what that something is makes all the difference.
    If the mass is released from the position in which it was attached, i.e. with the spring at its natural length, and you interpret the 10cm as the maximum extension that ensues, you get your 196N/m answer.
    If the mass is carefully lowered until the spring takes its weight, then you get the 98N/m answer.

    Edit: do you see why your second question needs the energy approach instead?
     
  12. Mar 2, 2016 #11
    It was mentioned before that if it means that 0.1 m is its new equilibrium Then its kinetic energy doesn't equal to 0.

    So instead of PE(gravity) = PE(Spring). It should be PE(gravity) = PE(Spring) + KE
    But that means you have to variables in one equations which is most of the time unsolvable and you don't want to use mg = kd
    How to get rid of the KE? Well at the maximum stretch there won't be kinetic energy so you can simply multiply 0.1*2 to give you 0.2 which is the maximum stretch.
    Now at that point you can use your PE(gravity) = PE(Spring)

    The question didn't really specify that but your 2nd question did. Notice what he said (Came to rest)
     
  13. Mar 2, 2016 #12
    There is a simple way of expressing @haruspex answer.
    • Case 1 : The block is released from the height where the spring is at its natural length. Then at equilibrium position, the block will have zero acceleration but nonzero velocity. Then the maximum elongation is not calculated as elongation at equilibrium, but as elongation when block has zero velocity. This requires an energy approach.
    • If the block is slowly brought to the equilibrium position, then it has both zero acceleration and zero velocity at the equilibrium position. Thus in this case, the weight of the block can be equated to the spring force to obtain maximum elongation.
     
  14. Mar 2, 2016 #13
    If you slowly bring it down, That means you have done work. Your work must be placed in the equation to get the right value of K
     
  15. Mar 2, 2016 #14
    I meant slowly as in with no KE. Zero acceleration. Then what you have added is nothing but the potential energy of the configuration.
     
  16. Mar 6, 2016 #15
    Ohhh yes.. I got it!!! Thank you all very much!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Spring constant
  1. Spring constant (Replies: 19)

  2. Spring constants (Replies: 6)

  3. Spring Constants (Replies: 3)

  4. Spring constant (Replies: 1)

Loading...