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Significant Figures with Kinetic Energy Formula?

  1. Jul 22, 2017 #1
    1. The problem statement, all variables and given/known data
    An object of mass m = 2.3±0.1 kg is moving at a speed of v = 1.25±0.03 m/s. Calculate the kinetic energy (K = 1 /2mv2 ) of the object. What is the uncertainty in K?

    I am not exactly sure if I used the error equation correctly when I start using Δ(v2). Could someone verify my logic here?

    2. Relevant equations
    K=0.5mv2

    Δ(v2)=nx^(n-1)(Δx)

    ΔK=(k )(z(Δm/m +(Δv2)/(v2))), k=0.5, z=1.8

    3. The attempt at a solution
    K=0.5mv2=0.5(2.3kg)(1.25m/s)2=1.8J

    Power Error Equation
    Δ(v2)=nx^(n-1)(Δx)
    =2(1.25)^1(0.03)

    Δ(v2)=0.075

    Multiplication and Constant Error Equation

    ΔK=(k )(z(Δm/m +(Δv2)/(v2)))

    K = 0.5 ⋅ 1.8 ⋅((0.1/2.3)+(0.075/1.252))

    K = 0.08
     
  2. jcsd
  3. Jul 22, 2017 #2

    haruspex

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    Where did the 0.5 come from?

    An easy way to check your answer is to plug in the values that would make K its max: 1.25+0.03, 2.3+0.1.
     
  4. Jul 27, 2017 #3
    Sorry this was meant to be K = 0.5 ⋅ 1.8 ⋅((0.1/2.3)+(0.075/1.252)), is that still correct?

    The 0.5 comes from the K=1/2 mv2

    I understand that the 0.5 is an exact number, but my thought is that the uncertainty that I am calculating here needs to be scaled ( or with the same ratio) after you put the uncertainty with mass and speed given in the beginning through the KE formula. OR does that not matter?
     
  5. Jul 27, 2017 #4

    I am using the exact constant rule: z = k Δx, whereas k has no uncertainty, or Δk=0. So I use 0.5 from the kinetic energy formula as k.
     
  6. Jul 27, 2017 #5

    SammyS

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    In general, Δ(xn) = nxn−1(Δx) .

    Specifically, Δ(v2) = 2v1(Δv) .
     
  7. Jul 27, 2017 #6

    haruspex

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    You already included that when calculating the 1.8J.
     
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