Classical Mechanics - Box sliding down a slope

[jinksys]

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?

 

[fluidistic]

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.

 

[richukuttan]

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,$\frac{d}{dt}(\dot x ^2 )=2(\dot x)(\ddot x)$

 

[richukuttan]

2) Why v0^2 and not v0 on the integral?

This is bcoz u r integrating [itex]{d}(\dot x²)[/itex] and not ${d}(\dot x)$.
So, the limits are 0 and v₀².

 

[PJC01]

1) The 2 in the second to last equation is likely a factor of 2 that comes from integrating the acceleration term in the equation. This is a common practice in classical mechanics to simplify equations and make them easier to solve.

2) The v0^2 in the integral represents the initial velocity squared, which is a result of the kinematic equation v^2 = v0^2 + 2ax. This equation is used to determine the final velocity of an object after traveling a distance x with an initial velocity v0 and acceleration a. Therefore, in the integral, the v0^2 term represents the initial velocity squared that is being integrated over the distance traveled.