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Unsure of ΔUspring in regards to ΔKE

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello everyone, I am having immense trouble with this problem. I'm not too sure if this is classified as Advanced Physics, but currently I'm a high school senior in AP Physics.

    You are given a spring with spring constant k=40,000 N/M. A 2 KG block is put onto the spring, compressing it 4 cm (.04 m). Find the velocity at half of the total height traveled. Find the total height.

    Let it be known that whenever I have an "i" next to a variable, it references initial in regards to said variable. Variables without an "i" are to be considered as the final versions of said variable.
    2. Relevant equations
    (kx^2)/2 - (kxi^2)/2
    (mv^2)/2 - (mvi^2)/2
    mgh-mghi
    Basically the Potential and Kinetic Energy equations.

    3. The attempt at a solution
    What I ended up doing was something to this degree.
    Given:
    K=40000
    m=2
    ΔX=.04
    Total height = (h+.04)

    ΔKE + ΔU + W' (No symbol for prime...) = 0
    W' = 0 because of no outside forces
    ΔKE = -ΔU
    (mv^2)/2 - (mvi^2)/2 = -[(kx^2)/2 - (kxi^2)/2 ]
    (mvi^2)/2 = 0 since vi = 0
    (kx^2)/2 = 0 since x = 0

    mv^2/2 = kxi^2/2
    mv^2 = kxi^2
    v= √(kxi^2/m)

    Well, it all went pretty downhill from here, but let's keep going...

    Given mv^2/2 - mvi^2/2 = -mgh + mghi
    mgh = 0 because h = 0
    -mvi^2/2 = 0 because vi = 0

    Since kxi^2/2 = mv^2/2
    I can substitute

    kxi^2/2 = mg((h+.04)/2)

    Plugging values in, I got

    40000(.04).04)/2 = 32

    2(10)(h+.4)/2 = 10h+.4
    32 = 10h+.4
    31.6=10h
    h=3.16 m

    Solving for v using newly found h, I did
    mv^2/2 = mgh/2
    v^2/2 = gh/2
    V^2 = gh
    V = √(31.6)
    V = ~5.2ish m/s

    My teacher doesn't mind discrepancies due to rounding, hence g = 10

    So, after speaking to my fellow classmates, I came to the conclusion I did something wrong. What though? Any help would be much appreciated. Tomorrow I'm talking to him about the problem to figure out my mistake. Or, maybe I'm just smarter than everyone else in my class and they got the wrong answer. We'll see.
     
  2. jcsd
  3. Nov 7, 2014 #2

    Simon Bridge

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    What makes you think something went wrong?
     
  4. Nov 7, 2014 #3

    ehild

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    These are not equations. The right sides are missing.

    From where do you count hight h? Apparently, you count it from the top of the uncompressed spring.
    How do you count gravity?

    You should include the gravitational potential energy, too.

    In principle, it is wrong, but gravity is small compared to the spring force, so your speed value is acceptable for the instant when the block leaves the spring.

    The velocity is not zero when the mass detaches from the spring.

    What do you mean with that equation? The velocity is not zero at half of the total height.
     
  5. Nov 8, 2014 #4
    It was late when I rewrote the problem from my mind.
    The right sides would simply be the ΔU or ΔKE in regards to the equation as stated above.
    The height would be the height, h, plus the compressed part of the spring. We are to assume h=0 when the spring is uncompressed. Sorry I forgot that bit of information.
    The speed of the block as it initially leaves the spring is not 0, my mistake.
    Substituting the spring potential energy equation with regards to kinetic energy, then subsequently substituting that equation with gravitational potential energy allows us to equate the two equations. The h we are using would be the total height, h, plus the compressed area of the spring, .04, all over two. That would be the middle height. The final equation, to the best of my very lackluster physics skills, does not imply velocity is 0 at h/2.
     
  6. Nov 8, 2014 #5

    ehild

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    You wrote that kxi^2/2 = mg((h+.04)/2).
    The left -hand side is the initial elastic energy, is it not? The right hand side is the potential energy at half the final height.
    If the velocity is not zero at half-height, there should be some kinetic energy included.
     
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