Combining Multiple Rules for Error Propagation

AI Thread Summary
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.
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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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