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Homework Help: Mechanics - Work and Kinetic Energy, (Spring on a Ramp)

  1. Jan 5, 2017 #1
    Hi everyone, I've been systematically working through the exercises in my first year college text book and have come up confused over this particular problem. I have a solution manual but it solves this problem differently than I did, though comes up with a similar answer. I've included my attempt at the problem. Hopefully someone can clear up my mistake if there is one. Much obliged.

    1. The problem statement, all variables and given/known data

    A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40 degrees above the horizontal. The glider has mass .09 kg. The spring has k= 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.8 m along the track before before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. a) What distance was the spring originally compressed? b) When the glider has traveled along the air track .8 m from its initial position against the compressed spring, is it still in contact with spring (MY Interpretation - What is the springs unstretched length but I can't be sure.)

    2. Relevant equations

    W = KE_f - KE_i = 1/2mv^2 - 1/2mv^2
    W = Fdcos(theta)
    which can simplify to
    W = mgh (in some instances I've encountered, such as projectile problems)
    W = 1/2kx^2 for a spring ( im not sure when I'm supposed to make it negative though...)
    3. The attempt at a solution

    I assumed that the question was implying that their was an initial speed to the spring (my solutions manual says there is no v_i , that it is equal to zero) the second it was released and that

    gravity would bring the velocity down to zero at it's highest point, so (1.8sin(40)) = h

    I figure that this is the point where I find out the energy imparted by the spring and equate it to the energy gravity "drains" away as it rises on the track.

    1/2kx^2 = mgh
    Where x is the compression distance and my unknown variable.
    .5(640)(x^2) = .09 (9.8)(1.8sin40)

    I get x =.0564 m

    My solutions manual says that I should have accounted for the work due to gravity somewhere ( which I thought I did by equating them together). It says that the x value is x = .0565 m (so I did something right).

    The solutions manual has a line that goes like this. I don't understand why it has cos130 in it.
    W_total = W spring + W weight = 0 (This I don't get, I thought their was an implied initial speed?)
    Shouldn't the above be equated to KE_f - KE_i ?
    W_w = mg(cos130)s the book uses s instead of height h, or x, or d
    W_spring = .09(9.8)(cos130)(1.8)

    I guess I'm going in circles, what I don't really understand is why there is a cos 130 in my book. The diagram shows its between the force mg and the incline.

    Maybe I'm looking for confirmation that I did the problem right, because I just don't understand how the solutions manual did it. It was pretty clear in most cases otherwise. This is chapter 6 and I've had only 2 or 3 problem solutions out of hundreds of them.

  2. jcsd
  3. Jan 5, 2017 #2


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    I think your solution is simpler. In general, there are two approaches: one is to have various energies positive and negative and the sum is always zero. The second approach is to have all energies as positive and use to context to decide when they are equal.

    Personally, unless the problem demands otherwise, I like to work with positive quantities. So, I would say:

    Potential Energy stored in spring: ##E_1 = \frac12 kx^2 \ ##, where ##x## is the compression of the spring.

    Final gravitational PE of mass: ##E_2 = mgh = mgl \sin \theta##

    And ##E_1 = E_2##

    Part b) seems to be answered already by part a).

    I have no idea why someone would use ##\cos(130)## instead of ##\sin(40)##
  4. Jan 5, 2017 #3


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    I think it takes the view that Wweight is the work done by gravity and the angle between the force of gravity and the displacement is 130 degrees.
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