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Mass and Spring Compression

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data

    (a) A block of mass m = 2.00 kg sits on a massless platform supported by a vertical
    spring with spring constant k = 55.0 N/m. How much is the spring compressed?

    (b) The block is now lifted off the platform and dropped from a height h = 1.5 m
    above the platform and spring. What is the speed of the block just as it reaches
    the spring?

    (c) What is the maximum compression of the spring by the block?

    (d) Just as the block leaves the spring moving upward, another identical block is
    dropped from a height h2 = 2.5 m above the equilibrium position of the spring.
    How far above the spring are the two blocks when they collide?


    2. Relevant equations

    F=ma
    F=-kx
    v2=u2 + 2as
    KE=0.5mv2
    PE=0.5kx2
    W=0.5kd2

    3. The attempt at a solution

    (a)
    I used ma=-kx and got 2(-9.8)=-55x →x=0.356m

    (b)
    Using v2=u2 + 2as, where s=-1.5, u=0, a=-9.8

    I got v=5.422ms-1

    (c)
    I used W=KE + PE

    And got d = 1.0935m

    (d)
    It is here where I have trouble. Could anyone tell me if I have used the right methods up til now? I get for part (d) I'm trying to find when both masses have equal h values but I don't know how to show that.
     
  2. jcsd
  3. Apr 30, 2014 #2
    The potential energy ## U= \frac{1}{2} kx^2 ## in the spring is now converted to kinetic energy ## K = \frac{1}{2}mv^2 ##

    Let ## s_1 ## represent the height of the block just about to leaving the spring, and ## s_2 ## be the height of the second identical block being dropped from rest.

    Thus at any time t,

    ## s_1 = v_0 t - \frac{1}{2}gt^2 ##

    Block on spring can be set at height = 0 (reference point). You have already solved ## v_0 = 5.6 m/s ## using mechanical conservation of energy. Since the block is thrown up it will start to decelerate immediately due to gravity.

    At any time t the position of the second block is given by,

    ## s_2 = h - \frac{1}{2}gt^2 ##

    Because the block starts at height ## s_2 = h_2 ## just sub that it for ##s_2##. We know the initial velocity is 0. The block once released is immediately accelerated downwards.

    At the collision point the position of both blocks are the same i.e ## s_1 = s_2 ##

    solve for t and sub this back into either ## y ## equation.
     
  4. Apr 30, 2014 #3
    Everything you've done up to the last part is correct.

    The ending can be viewed in a couple ways, but the way that I think is probably the least work intensive is to look at the relative velocities of these objects.

    If you were to compare the relative velocities, you could simplify this problem into one block going at a single velocity toward a static point. The relative velocity of the two objects would just be the difference of the two velocities. Well, since you know that they both have a constant downward acceleration, the acceleration aspect would end up cancelling out. The second block was dropped from rest, so the system can be simplified down to the single block's initial velocity going upward, and the time it down take for it to travel a distance of 2.5 meters.

    Once you have this time, plug it back into either one of the motion equations for your original system of two blocks, and this will give you the distance at which they meet.
     
  5. Apr 30, 2014 #4
    Right so using the first equation I got:

    0 = 5.6t - 0.5(-9.8)t2

    Getting t on its own for t=1.1s

    I then subbed that in to the second equation for:

    2.5 = h - 0.5(-9.8)(1.12)

    Getting h on its own for h=3.429m which doesnt make sense so I must be going wrong somewhere. Thanks for the help by the way.
     
  6. Apr 30, 2014 #5
    Why did you set that to zero? That means you're solving for the time it would take to travel 0 meters, which doesn't make a ton of sense.
     
  7. Apr 30, 2014 #6
    I said the block was at height h = 0 not ##s_1 = 0 ## recall that ## y = x_0 + v_0t + \frac{1}{2} at^2 ## so ## x_0 = 0 ## for block on spring.
     
  8. Apr 30, 2014 #7
    ## y_1 = v_0t - \frac{1}{2}gt^2 = h-\frac{1}{2}gt^2 = y_2 ##

    ## t = \frac{h}{v_0} ##

    ## y_1 = h - \frac{1}{2}g \Big(\frac{h}{v_0}\Big)^2 ##

    ## h = 2.5 m ##
     
  9. Feb 9, 2016 #8
    I have just studied the whole thing of springs and Energy few hours ago. But I thought I could share my idea.

    I have got the velocity that the object on the string will launch off towards the second object. After all, The work that the spring does on the object will be equal to its kinetic energy.. I guess what you have a problem in is determining where they will meet.

    Take a simple situation, two objects with a constant speed of v1 and v2.
    The first object equation is: df = di + v1*t
    Lets assume that di = 0
    and the second object that is falling from a height lets just say h:
    df = h - v2*t
    Look at the minus there because it is going in the opposite direction.
    What I want is the time that the two objects will collide on. You just have to put 1 into 2 and you will get your answer.
    The same thing applies to the spring situation but you dont need to put the acceleration as rellek said you will just end up removing it.
     
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