Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider a block sliding down an incline plane at an angle $\theta$ with the horizontal. For the acceleration as a function of $\theta$ I find $$\ddot{x}=g \ \sin\theta $$ My text then claims we can find the block's velocity after it moves a distance $x_0$ from rest by multiplying both sides by $2\dot{x}$ and doing the following:

$$2\dot{x}\ddot{x}=2\dot{x}g \ \sin \theta$$ $$\frac{d}{dt}(\dot{x}^2)=2g \sin\theta\frac{dx}{dt}$$ $$\int_0^{v_0^2}d(\dot{x}^2)=2g\sin\theta\int_0^{x_0}dx$$ $${v_0}^2=2g\sin\theta \ x_0$$ $$v_0=\sqrt{2g\sin\theta \ x_0}$$

I think I understand up until the 3rd line. The $dt$'s disappear because both sides are exact differentials, yes? Then in the next step, why does $\dot{x}$ (the velocity) vary from 0 to ${v_0}^2$? Thanks in advance.

Also, is there another way to do this?

share|improve this question
3  
If $\left.\dot{x}\right|_{x=x_0}=v_0$ then $\left.\dot{x}^2\right|_{x=x_0}=v_0^2$, no? –  Kyle Kanos Oct 8 '13 at 13:00
    
I will fill in the missing step. In the second line, he has $\frac{d}{dt}(\dot{x}^2)=2g \sin\theta\frac{dx}{dt}$, then integrates w.r.t. $t$ to get $\int_{t_i}^{t_f}\frac{d}{dt}(\dot{x}^2) dt=2g\sin\theta\int_{t_i}^{t_f}\frac{dx}{dt} dt$. Then he does $u$-substitution on both sides. On the left $u$ is $\dot{x}^2$ and on the right $u$ is $x$. This gives you his third line. If anyone wants to turn this into a complete answer, go ahead. –  NowIGetToLearnWhatAHeadIs Nov 8 '13 at 0:09

2 Answers 2

A simpler way to this is to use the conservation of energy. If surface is frictionless, Then P.E = K.E ($mg x_o sin \theta= 1/2 m v_o^2$), this would lead to the same final answer).

share|improve this answer

When you go from

$\frac{d}{dt}(\dot{x}^2)=2g \sin\theta\frac{dx}{dt}$

to the next step, the rigorous way to think about it is to integrate wrt $t$,

$\int_{t_0}^{t_1}\!dt\, \frac{df}{dt} = \int_{f(t_0)}^{f(t_1)} df$.

In this case, $f(t) = \dot x$, and $f(t_1) = v_0$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.