Derivative Method for Error in Kinetic Energy formula

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Homework Statement



Finding error in kinetic energy


Homework Equations



K = [itex]\frac{1}{2}[/itex] m v2


The Attempt at a Solution



Measured mass and velocities have errors in them. So we have to use derivative method to calculate uncertainty in KE which is to find the square root of the derivatives of K with respect to m and v and multiply by the errors of the variable you took the derivative of.

[1] derivative of K with respect to m = 1/2 v2 times error in the mass
[2] derivative of K with respect to v = m v times error in the velocity

to calculate the error in K we have to take the square root of the addition of the square of [1] and square of [2].

What I don't understand is why the units don't match with equation [2]. Units should be kg2m2/s2
 
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vela
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Homework Statement



Finding error in kinetic energy


Homework Equations



K = [itex]\frac{1}{2}[/itex] m v2


The Attempt at a Solution



Measured mass and velocities have errors in them. So we have to use derivative method to calculate uncertainty in KE which is to find the square root of the derivatives of K with respect to m and v and multiply by the errors of the variable you took the derivative of.

[1] derivative of K with respect to m = 1/2 v2 times error in the mass
[2] derivative of K with respect to v = m v times error in the velocity

to calculate the error in K we have to take the square root of the addition of the square of [1] and square of [2].

What I don't understand is why the units don't match with equation [2]. Units should be kg2m2/s2
[1] should be
$$\frac{\partial K}{\partial m} = \frac{1}{2}v^2$$ and
$$\Delta K = \frac{\partial K}{\partial m} \Delta m.$$ Perhaps that's what you meant, but what you wrote is
$$\frac{\partial K}{\partial m} = \frac{1}{2}v^2 \Delta m.$$ In any case, why do you think the units aren't working out?
 

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