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!

Hooke's law

  1. Nov 27, 2004 #1
    Hi
    Could you please help me solve these problems?


    A skater with mass m = 80 kg and speed v = 8 m/s hits (colides with ) a spring with spring constant 40 N/m.
    a .What is the maximum compression of the spring or in other words , how far will the sping be strecthed?

    b. What is the speed of the skater after the spring has been compressed 5 m?



    The same spring is now hung vertically from a tower on a bridge such that the free end is at the same level with the road , the skater hangs on to the free end of the spring and jumps over the bridge with an initial velocity v = 0 .

    c How far is the spring extended when the skater has attained maximum speed?
    d. What energy transfers occur when the skater falls towards the water?

    Calculate the following :
    1. The skater's maximum speed during the fall
    2. How far under the road does the skater stops?
    3 The geatest force the skater will experience form the spring during the fall.


    Thanks for your help and guidiance in advance.
     
  2. jcsd
  3. Nov 27, 2004 #2
    Hint:
    For the first problem use the formula for potential energy in the spring


    [tex] E_p = \frac{1}{2} k x^2 [/tex]

    x is the distance from the equilibrium point
     
    Last edited: Nov 27, 2004
  4. Nov 27, 2004 #3

    Doc Al

    User Avatar

    Staff: Mentor

  5. Nov 27, 2004 #4
    yeah..for the first problem...use the fact that KE = 1/2 mv^2....and U = 1/2 kx^2.

    (If this problem had a mass attached to the spring, you would have to consider the conservation of momentum; however, in this case, you do not need to, because the skater is colliding with a spring that has no mass attached to it. Thus, the skater retains the same momentum and thus the same speed. Therefore, you can simply proceed to use the conservation of energy.)
     
  6. Nov 27, 2004 #5
    for the second problem..also use the conservation of energy (you can do this, because there are no inelastic collisions)....

    so...basically..use KE = 1/2 mv^2....and U of a spring = 1/2 kx^2...

    ...also...Gravitational Potential Energy = mgh
     
  7. Nov 27, 2004 #6
    oh...sorry..didn't realize that this thread had already been posted elsewhere
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Hooke's law
  1. Hookes law (Replies: 1)

  2. Hookes Law (Replies: 17)

  3. Hooke's law (Replies: 7)

  4. Hooke's Law and force (Replies: 4)

  5. Hooke's law (Replies: 11)

Loading...