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Homework Help: Energy Problem

  1. Mar 21, 2007 #1
    1. The problem statement, all variables and given/known data

    "A spring of negligible mass and spring constant 60 N/m is attached to a block of mass 5 kg, which rides on a horizontal surface with a frictional kinetic coefficient of 0.15. A string then goes over a perfect pulley ( no friction or moment of inertia) and attaches to a hanging block of mass 2 kg. Initially, the spring is at the unstretched and uncompressed position, and the blocks are released at rest."

    Ok... they first want me to find the a) fricitional force acting on the 5 kg block as slides to the right

    and b) then the maximum extension of the spring

    3. The attempt at a solution

    Ok.. What I really want is for someone to agree with the logic that I will state below in trying to solve these problems and if not, please correct me.

    a) Ok.... So the only way I think I can solve this is through Netwon's second law right?

    So for Block attached to spring, Fy= Fn-m1g=0
    and Fx=T-f=m1a? Right?... if we consider the axes as they regularly are?
    For Hanging Block.... Fx=m2g-T=m2a... if we treat the x as positive pointing down...
    .... but if this is solve in this way, then how do I find the acceleration?... I mean I can cancel the two tensions.... Fx=m2g-f=(m1+m2)a... but how do I get the acceleration...

    Or is it that simple... that I just do Fn=m1g and mutiple the frictional kinetic coeffiecient?

    Or the second part...

    I know we have to use the conservation of energy....

    So Ei=Ef

    That means, initially there is no kinetic energy or potentinal energy at rest

    so... Ei=0???

    so... Ef= 1/2kx^2+1/2(m1+m2)v^2+m2g(-y)+f(-y)=0..Right????

    I mean the spring has potential energy, then the two blocks are moving at the same velocity as a system, so they have kinetic energy, and the hanging block moves down, so it has potential energy, and the frictional force disippates energy in the form of thermal energy? Is this right... and if so.... how do we factor in the unknown factors of the velocity of the system, and the distance the two blocks move?

    Thank you.
  2. jcsd
  3. Mar 21, 2007 #2


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    Part 'a' just asks you to calculate the friction force, f, which you have noted is as simple as f = u_k(F_n). Then you can solve for T using the 2 equations you have correctly identified, and once you solve for T, you can get the spring extension using Hooke's law without having to use the conservation of energy approach.
  4. Mar 21, 2007 #3
    Wait... so... How can you use Hooke's Law, once you have the tension, to find the maximum extension?



    ... But even if I wanted to do it the other, way, though harder, would previouis energy equations be correct? or wrong? Thanks.
  5. Mar 21, 2007 #4
  6. Mar 21, 2007 #5
    Question: Are the two masses attached by a spring or by a string?
  7. Mar 21, 2007 #6
    The two masses are attached by a string, but one of the masses is held in place by a spring unstretched and uncompressed....
  8. Mar 21, 2007 #7
    Ah, OK. You said in your first post that:

    which is true before the blocks are released from rest. However, note that f is the force of static friction which you are not asked to find. You are asked to find the force of kinetic friction. This, as PhanthomJay pointed out, is just the product of the coefficient of kinectic friction and the normal force on the mass.

    To start solving part b, first determine the net force in the horizontal direction of the mass attached to the spring as the mass is sliding. Once you have this, ask yourself how you can determine the maximum extension of the spring from it.
  9. Mar 21, 2007 #8

    .. So you're saying that I should find the acceleration of the system, then use it to find the tension in both ropes, to find the maximum extension??

    So.... T=-kx???

    But what if I wanted to do it using the conservation of energy???
  10. Mar 21, 2007 #9
    Both ropes? What ropes? I thought there was only a spring and a string. Anyways, read what I said again.

    No. Draw yourself a free body diagram of the situation. Analyze the forces and you should quickly see that what you wrote above is wrong.

    I'm not sure that would help, although I may be wrong.
  11. Mar 21, 2007 #10
    ..... Ok, so I see that for block #1 held in place by the spring,

    Fx= T-f(kinetic friction)=m1a
    Fy= Fn-m1g=0

    .. for block #2..... m2g-T=m2a... but I still don't understand how you can get the extension, without conservation of energy????
  12. Mar 21, 2007 #11
    Aren't you forgetting something, viz. the force by the spring?
  13. Mar 21, 2007 #12
    Oh.. so is it.... Fsp+T-f=m1a (box#1)
    mg-T=m2a (box #2)......... But where would I go from there, since the acceleration, Fsp, and T are unknown values? Thank you.
  14. Mar 21, 2007 #13
    Question: In what direction is Fsp acting? Do you know about Hooke's law?
  15. Mar 21, 2007 #14
    Hooke's law, F=-kx, so the Force of the spring is acting opposite to the tension of the rope? Right.. missing something? So, assuming this is right,...


    ... still if we don't know Fspring, tension, and acceleration are????
  16. Mar 21, 2007 #15
    Good, except that your second equation should be m2g-T=m2a. You know what the acceleration is, you just have not realized it yet. (Think about the system when the spring is in its new equilibrium position.)
  17. Mar 21, 2007 #16
    So... Hooke's law, says F=-kx=ma=But, since mass is neglected,

    does, a=-kx?????


    T-Fsp-f=m1a, so

  18. Mar 21, 2007 #17
    No no no. I guess I'll have to be more explicit: Consider the system when the masses are at rest again. What is the acceleration of the masses? What does the value of x represent when the masses are at rest again? Be careful with the sign in Hooke's law.
  19. Mar 21, 2007 #18

    x represents the distance block #1 travels horizontally and x represents the distance block #2 travels vertically down, so are we trying to apply kinematics?, ....

    v^2/x=a??? I really can't think of anything else as of the moment??
  20. Mar 21, 2007 #19
    Yes, that is true. I was hoping you would say something about x as it pertains to Hooke's law.
  21. Mar 21, 2007 #20
    ... x is how far the spring is stretched/extended... where can I go from here?
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