Work, Velocity & Force vectors Help

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SUMMARY

The discussion focuses on the application of the work-energy theorem in Classical Mechanics, specifically the equation F . dr = (MV2^2)/2 - (MV1^2)/2. Participants clarify the origin of the number 2 in the denominator, which arises from applying the product rule to the derivative of velocity squared. The correct interpretation involves recognizing that d/dt(x^2) = 2x(dx/dt), linking the velocity vector to the derivative. This understanding is crucial for grasping the mathematical foundations of physics equations.

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Jonnyb42
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This is probably more of a math question, but it pertains to physics.

Today in Classical Mechanics the professor was showing the following,

F . dr = (MV2^2)/2 - (MV1^2)/2

but in great length, one of the steps along the way I did not understand how he did it... it was this:

screw latex, if someone could please post a link to somewhere on the forums that shows how to use latex properly, I can't use it to save my life...
here is the step with good old microsoft:

[PLAIN]http://mynqa.com/Cargo/unk.bmp

My questions are, where did the number 2 come from in the denominator? Also, how did he bring in the velocity vector into the derivative like that?

Thanks for any help.
 
Last edited by a moderator:
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Jonnyb42 said:
screw latex, if someone could please post a link to somewhere on the forums that shows how to use latex properly, I can't use it to save my life...

See my sig :wink:

here is the step with good old microsoft:

[PLAIN]http://mynqa.com/Cargo/unk.bmp

My questions are, where did the number 2 come from in the denominator? Also, how did he bring in the velocity vector into the derivative like that?

Thanks for any help.

Just apply the product rule to the right-hand side of the equation...
 
Last edited by a moderator:
It is essentially the vector equivalent of the derivative of a square.

d/dt(x2)=2xdx/dt
 

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