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Hooke's Law and a Can -- What is the speed and acceleration?

  1. Dec 2, 2015 #1
    1. The problem statement, all variables and given/known data
    One end of a horizontal spring with force constant (76.0 N/m) is attached to a vertical post. A 4.00-kg can of beans is attached to the other end. The spring is initially neither stretched nor compressed. A constant horizontal force of 57.0 N is then applied to the can, in the direction away from the post.

    Part A
    What is the speed of the can when the spring is stretched 0.400 m?

    Part B
    At the instant the spring is stretched 0.400 m, what is the magnitude of the acceleration of the block?

    2. Relevant equations
    Us = 1/2kx^2
    Ke= 1/2mv^2

    3. The attempt at a solution

    I considered setting

    1/2kx^2 =1/2mv^2 and solving for "v". I got the answer 1.74m/s which is incorrect. As for part B I considered using the equation F =ma.
    57N = 4kg(a) a = 57/4 = 14.25 m/s^2. However that was incorrect as well.

    Please assist! I have no idea where to go from here.
     
  2. jcsd
  3. Dec 2, 2015 #2
    Ask yourself, is the can at maximum stretch when it is at 0.4 m? That is the interpretation I get from your energy relationship.
     
  4. Dec 2, 2015 #3
    I do not think it is at maximum stretch at .4m. I think it can go further. But I am not sure as how to proceed with this.
     
  5. Dec 2, 2015 #4
    Is there a way you can determine the maximum stretch?
     
  6. Dec 2, 2015 #5
    I think that it is 1/2kx^2. But because it is stretched would it be negative?
     
  7. Dec 2, 2015 #6
    That's the speed when the kinetic energy equals the potential energy, but why would you assume those two would be equal? Or were you just hoping it would work and not making that assumption at all?

    That would be the acceleration of the can if only the 57 N force is applied, but that's the case only when the spring is applying no force.

    You haven't really tried to do anything based on anything about the spring and the can. You're just grabbing equations and plugging in numbers. You need to make some assumptions about how 57 N forces, cans, and springs behave. Have you made a sketch and thought about how the forces affect the can's motion? Or alternatively thought about the amount of energy in the system and how it's changing due to the application of the 57 N force?

    You have to meet us half way here.
     
  8. Dec 2, 2015 #7
    If you use energy than you will have two unknowns to worry about. Why don't you use force? Will you be able to stretch this spring forever with a constant force of 57N? What are the limits of the spring?
     
  9. Dec 2, 2015 #8
    Eventually, the spring will need to return back to the equilibrium point where x = 0. Otherwise the object will be deformed and won't be able to return back to it's original state.
     
  10. Dec 2, 2015 #9
    your post title states Hooke's Law. Can you use that somehow?
     
  11. Dec 2, 2015 #10
    Serela: You should follow Mister T's suggestion, and draw a free body diagram. Or do you feel you have advanced beyond the point where you need to use free body diagrams? What are the horizontal forces acting on the can? What is the force balance on the can associated with the action of these forces?

    Chet
     
  12. Dec 2, 2015 #11
    Okay, I did draw the diagram as Mister T suggested. I noticed the change in position, because of the force applied the initial conditions (x =0) do not work... So I would need to solve for the new x which will become the x initial, which I believe that is what you, rmthomps, was hinting at.

    When I did that I took the Hooke's equation F=kx and just solved for x so, F/k = x. Which I have both values. Because the initial conditions do not apply, I would not be able to equal 1/2kx^2 =1/2mv^2 which is what Mister T noted. Thus the "new" equation would be, "1/2kxinital^2" =1/2kxf^2 +1/2mvf^2.

    And thus solve for vf.

    For the acceleration equation, would I be able to just use vf^2 =vo^2 +2adeltax?
     
  13. Dec 2, 2015 #12
    I still don't see a force balance equation on the can. And I still don't see a correct energy balance. The force balance can be used to derive the energy balance, by you have got to get the force balance correct first. Didn't they teach you how to do force balances in your course?

    What is the force that the spring exerts on the can, and, in what direction? What is the external force applied to the can, and, in what direction? What is the net force applied to the can?

    Chet
     
  14. Dec 2, 2015 #13
    Opps! The force the spring exerts on the can is leftward which I chose to be negative because the spring is trying to return back to it's on. The external force is in the rightward direction which I chose to be positive.
     
  15. Dec 2, 2015 #14
    So, let's see your force balance equation already.

    Chet
     
  16. Dec 2, 2015 #15
    -F spring exerts + Fexternal = 0 because afterwards they no longer exert a force on the object.
     
  17. Dec 2, 2015 #16
    I don't see ma = m dv/dt in your equation, and I don't see the spring force expressed in terms of k and x.
     
  18. Dec 2, 2015 #17
    -F(spring) +F(ext) = ma
    -F(spring) +F(ext) = m dv/dt
    integrate
    -F(spring)x + F(ext)x = mv
    v= -fspringx +f(ext)x/m
     
  19. Dec 2, 2015 #18
    What about the effect of the applied force of 57 N? That's got to make a difference in the way the can moves.

    But you're not interested in what happens then! You're interested in what happens when x = 0.4 m. If in fact the net force were zero, the acceleration would be zero, and you'd have you're answer!
     
  20. Dec 2, 2015 #19
    v = -kx^2 + Fext (x) / m
     
  21. Dec 2, 2015 #20
    This is not working, either. You can't integrate a variable force like the spring force in the same way that you integrate a constant force like the 57 N force. You can see that the units don't match in your final result!
     
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