Classical Mechanics - Box sliding down a slope

AI Thread Summary
The discussion focuses on two questions regarding equations from Thorton's Classical Dynamics book related to a box sliding down a slope. The first question addresses the placement of the factor of 2 in the equation for the derivative of velocity squared, which is explained by the chain rule of differentiation. The second question concerns the use of v0^2 in the integral instead of v0, clarified by noting that the integration is performed on the differential of velocity squared. The participants confirm that the limits of integration are 0 and v0^2 due to this differentiation. Understanding these concepts is crucial for grasping the dynamics of motion on slopes.
jinksys
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I'm on pg 56 of Thorton's Classical Dynamics book and I see this: Imgur Link

Two questions: 1) Where does the 2 go on the second to last equation. 2) Why v0^2 and not v0 on the integral?
 
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jinksys said:
I'm on pg 56 of Thorton's Classical Dynamics book and I see this: Imgur Link

Two questions: 1) Where does the 2 go on the second to last equation. 2) Why v0^2 and not v0 on the integral?
1)\frac{d}{dt}(\dot x ^2 )=2\dot x \ddot x.
2)I'm guessing it's because of the differential which is a differential of velocity squared.
 
fluidistic said:
1)\frac{d}{dt}(\dot x ^2 )=2\dot x \ddot x.
This is bcoz
\frac{d}{dt}(\dot x ^2 )=\frac{d}{d\dot x}(\dot x ^2 )*\frac{d}{dt}(\dot x)
So,\frac{d}{dt}(\dot x ^2 )=2(\dot x)(\ddot x)
 
jinksys said:
2) Why v0^2 and not v0 on the integral?
This is bcoz u r integrating {d}(\dot x<sup>2</sup>) and not {d}(\dot x).
So, the limits are 0 and v02.
 
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