# Uncertainty of Kinetic Energy

## Homework Statement

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 =(1/2)mv2) of the object. What is the uncertainty
in K?

## Homework Equations

Δz = |z| ( Δx/x + Δy/y ) - Multiplication

Δz = n (xn-1) Δx - Power

## The Attempt at a Solution

k=1/2 mv2

(power)
Δv = 2(1.25)1(0.03)
=0.075

(multiplication)
Δk = k(1/2) (0.1/2.30 + 0.075/1.25)
=0.09

1.80 ± 0.09 kg*m2/s2

Just wondering if this is correct or if I have gone about this wrong. I'm I correct in multiplying K by 1/2 in the last step?

## Answers and Replies

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haruspex
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Δv = 2(1.25)1(0.03)
=0.075
More accurately, that's Δ(v2).
(multiplication)
Δk = k(1/2) (0.1/2.30 + 0.075/1.25)
How do you justify the inclusion of the factor 1/2? I don't see that in the equations you quoted. Let's say we replace the 1/2 with an unknown, a:
k = a m v2
Δk/k = Δa/a + Δm/m + Δ(v2)/(v2)
Knowing that a = 1/2, no error, what would you write for Δa?

Okay I did the work right though I just wrote delta V instead of delta v^2?

1/2 is just a constant in the kinetic energy formula. I just treat it as something still being multiplied?

haruspex
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Okay I did the work right though I just wrote delta V instead of delta v^2?
Yes.
1/2 is just a constant in the kinetic energy formula. I just treat it as something still being multiplied?
No. Think about what I wrote before. 1/2 is a precisely known constant. If z = xy and x is precisely known, what do you get for Δz/z?

Δz/z = Δy/y ?

haruspex
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Δz/z = Δy/y ?
Quite so. Note that x has disappeared. In the present context, x represents the factor 1/2.

rude man
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I guess they're not teaching standard deviation error propagation any more these days? because those formulas are substantially different from the "simplified" ones given by the OP.

A good basic treatment is at http://www.rit.edu/~w-uphysi/uncertainties/Uncertaintiespart2.html [Broken]

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haruspex
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Gold Member
I guess they're not teaching standard deviation error propagation any more these days? because those formulas are substantially different from the "simplified" ones given by the OP.

A good basic treatment is at http://www.rit.edu/~w-uphysi/uncertainties/Uncertaintiespart2.html [Broken]
SDE propagation is not necessarily appropriate.
The link you posted isn't bad, but it fails to consider a few things.
1. What it classes as random errors includes some which are repeatable. If I measure a length to the nearest mm by eye, I cannot really judge the fractions of mm, so if the actual length is 19.294... mm it doesn't matter how often I measure it I will get 19mm. So the error has a flat distribution of ±0.5mm. This complicates SDE analysis a little. (The link mentions 'least count' but does not properly consider the consequences.)
2. If two engineering parts have specs of ±1mm, and the design requires that their sum must be under some value, the engineer will quite rightly calculate the total uncertainty at 2mm. If the plane falls out of the sky she can't blame it on a statistical fluke.

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rude man
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If you do worst-case design as opposed to std deviation, in most cases you'd be out of business in a hurry.

Yet in certain situations like aircraft safety, yes, you need to do wcd.

Blake_ap1
Relevant equations

Δz = |z| ( Δx/x + Δy/y)

Δz = n (xn-1) Δx

k=1/2 mv2 k = 1.796875

Δ(v2) = 0.075

Δk = k(0.1/2.30 + 0.075/1.25)

= 1.8 ( 0.103) = 0.186

= 1.8 ± 0.186 J

Is this correct?

haruspex
Homework Helper
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0.075/1.25)
That term seems to represent Δ(v2)/v. Did you mean that?

Blake_ap1
That term seems to represent Δ(v2)/v. Did you mean that?
Yes

haruspex