Combining Multiple Rules for Error Propagation

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To calculate the kinetic energy of an object with mass m=2.3±0.1kg and speed v=1.25±0.03m/s, the formula K=1/2mv² is used, resulting in K=1.8 J. The uncertainty in kinetic energy requires applying the product rule for error propagation, which states that for a product z=xy, the relative uncertainty is δz/z=δx/x+δy/y. For uncorrelated variables, uncertainties should be combined in quadrature, meaning δz²=δx²+δy². The discussion emphasizes the importance of correctly applying these rules to estimate the uncertainty in kinetic energy accurately. Understanding these error propagation methods is crucial for precise calculations in physics.
ELLE_AW
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Homework Statement


  1. An object of mass m=2.3±0.1kg moves at a speed of v=1.25±0.03m/s. Calculate the kinetic energy (K=1/2mv2) of the object and estimate the uncertainty δK?

Homework Equations


- Addition error propagation--> z = x + y and the Limit error--> δz = δx + δy

- Exponent error propagation --> z = xn and the Limit Error --> δz = nxn-1(δx)

- K = 1/2mv2

The Attempt at a Solution



This is what I attempted, but I really don't think it's right. I basically just included the exponent error propagation, but how does the multiplication of mv2 get incorporated?

- K = ½ mv2 = ½ (2.3kg)(1.25m/s)2 = 1.7969 kg m2 s-2 = 1.8 J

- Uncertainty of K = (m)2v1(δv) = (2.3)(2)(1.25)(0.03) = 0.1725 = 0.17
How do I combine these two rules when calculating the uncertainty of the kinetic energy?

 
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ELLE_AW said:
Relevant equations
You seem to be missing the product rule:
If z=xy then δz/z=δx/x+δy/y
 
ELLE_AW said:
Addition error propagation--> z = x + y and the Limit error--> δz = δx + δy
This is error propagation for correlated variables x and y. If your variables are uncorrelated, they should be added in quadrature, i.e.,
$$
\delta z^2 = \delta x^2 + \delta y^2.
$$
The same is true for uncorrelated relative errors in the case of a product.
 
Got it, thank you!
 
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