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Question on Hooke's Law

  1. Sep 25, 2005 #1
    Hello,

    I found one of my this homework question difficult and was wondering if anyone could please help me.

    A 180g trolley is placed on frictionless air track. One end of the trolley is attached to a spring of spring constant 50 N/m. The trolley is pushed against a fixed support so that the compression of the spring is 8 cm. The trolley is then released.

    (a) What is the initial acceleration of the trolley when it is released?
    (b) What is the initial energy stored in the spring?
    (c) Calculate the final speed of the trolley along the frictionless track. You may assume that there is 100% transfer of energy from the spring to the trolley.

    Any help would greatly be appreciated.



    charikaar
     
  2. jcsd
  3. Sep 25, 2005 #2

    Päällikkö

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    (a) F = ma
    (b) [tex]W = \frac{1}{2}kx^2[/tex] You can obtain that by integrating, but I'm not sure if you're familiar with that stuff.
    (c) Energy principle
     
  4. Sep 25, 2005 #3
    hello,

    Thanks for the quick reply.


    Mass of trolley: 180g or 0.18 kg

    compression: 8cm or 0.08m

    Using F=extentionxconstant to find Force
    50/0.08=625N

    Now using F=ma to find acceleration

    625/0.18=3472.22....Am i correct?

    (b) using W=1/2kx^2

    25x0.64=16J

    (c) I've no idea how to start this one.


    thank you.


    Charikaar
     
  5. Sep 25, 2005 #4

    Päällikkö

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    (a) Instead of calculating the force right in the beginning, solve the problem with the letters, and into the final equation, plug in the requested stuff. Oh, and F = -kx, not -k/x
    (b) Are you using SI-units? I got a different answer.
    (c) What sort of energy does the potential energy of the spring transform into?
     
    Last edited: Sep 25, 2005
  6. Sep 26, 2005 #5
    Hi again,

    I've just started AS physics and am not familiar with this stuff much. I don't know where does my teacher get these question from when they are not in the book.


    (a) k=50 N/m, x=0.08 m using F=kx, F=50x0.08=4N

    Now using F=ma to acceleration
    a=F/m, 4/0.18=22.22....N/mkg


    (b) This is AS Physics question.
    Elastic Potential Energy=1/2xstretching forcexextension
    =1/2x8x0.08=0.32J




    (c)I think it transfers to energy of movement. Could you please give me equation for the solution.


    Thanks for your help.


    Charikaar
     
  7. Sep 26, 2005 #6

    Päällikkö

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    (a) Correct. From F = ma, we can see that the unit of F is [itex]kgm/s^2[/itex] This divided by kg, we get [itex]m/s^2[/itex], which is the unit of acceleration.
    (b) My answer's half that.
    (c) Kinetic energy: [tex]\frac{1}{2}mv^2[/tex].
     
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