1. Not finding help here? Sign up for a free 30min 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 question

  1. May 3, 2005 #1
    Hey I've worked on this question for so long and can't get the right answer (I don't know what it is though).

    A spring (k = 600 N/m) is placed in a vertical position with its lower end supported by a horizontal surface. The upper end is depressed 20 cm, and a 4.0 kg block is placed on the depressed spring. The system is then released from rest. How far above the point of release will the block rise?
    State your answer in centimetres to 2 significant figures.

    I've gotten the answer of 27cm by two different methods but my online tutorial still says it's wrong. Firstly, I used 1/2*k*x^2 where x=0.2 to get the total mechanical energy for the system (12J). At the highest point, the velocity of the block will be 0, therefore its kinetic energy will be 0. Therefore, the energy in the spring + the gravitational potential energy will = 12J. Using this i got x=7cm which means the block rose 27cm.

    I then used the equation that Fnet=Fspring-Fweight
    ma=-600x-mg
    dv/dt=-150x-9.8
    v=-75x^2-9.8x+C
    subbed in @ x=-0.2, v=0
    :. v=-75x^2-9.8x+1.04

    Using this i got v=0 at -20cm and 7cm, which gives the same answer as above. What am I doing wrong?
    Thanks
     
  2. jcsd
  3. May 3, 2005 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    One more way to do this problem is to find the equilibrium position of the mass on the spring, and find the initial position relative to that. You know that if the equilibrium position is above the starting point, the initial displacement from equilibrium will be the amplitude of the motion, and the mass will move twice that far. I think you will still get your answer.
     
  4. May 3, 2005 #3

    minger

    User Avatar
    Science Advisor

    Yes, I agree with OlderDan. It seems you are doing it much more complicated than it has to be. The system consisting of a the spring and mass will reach equilibrium when forces are equal. That means that the weight of the mass will be equal to the force of the spring pushing upwards. If you have the mass of the block, then you have the weight, then solve the spring force equation for x. Subtract from your intial starting point to get the rise.
     
  5. May 3, 2005 #4
    I tried doing that too. Is this what you mean
    -kx = mg
    x=mg/-k
    using this i get x=-7cm, which would mean the block moves up 13cm right? This was my initial answer to the problem but that was wrong too.
     
  6. May 3, 2005 #5

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Yes. -7cm means you are starting 13cm below and you will go 13 cm above so you move 26 cm. It's the same answer you keep getting within the round off. Either you have interpreted the problem wrong, or the answer you have been given is wrong. It happens now and then.
     
  7. May 3, 2005 #6
    What I"ve put above is the question word for word from the online tutorial. I don't think the correct answer according to the online tut would be wrong as this is an assessment item. Is there any other way this question can be interpretted?
     
  8. May 4, 2005 #7

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Yes there is. Sorry for not catching it sooner. The 4 kg mass is PLACED on top of the spring, not attached to the spring, as I had assumed when thinking about this earlier. That makes a whole lot of difference.
     
  9. May 5, 2005 #8
    can you please elaborate on how this would affect the calculations because I don't see it? thanks
     
  10. May 5, 2005 #9

    minger

    User Avatar
    Science Advisor

    Is there an answer that is around 7? To two significant figures, I came up with 6.54 cm. It asked how far above the initial point will the block rise, so you can ignore any negative signs or adding to initial value of displacement

    I would think if 6.54cm doesn't work, then somethings wrong.
     
  11. May 5, 2005 #10

    Doc Al

    User Avatar

    Staff: Mentor

    I'm sure they want you to assume that all of the initial spring potential energy gets transformed to gravitational potential energy.
     
  12. May 5, 2005 #11
    lol of course. That gave the right answer. thanks so much for everyone's help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Spring question
  1. Rocket Spring question (Replies: 4)

  2. Spring pull question (Replies: 1)

  3. Spring questions (Replies: 4)

  4. Spring Questions (Replies: 1)

Loading...