Finding position wit the energy method

[anubis01]

Finding position with the energy method

Homework Statement

A block slides on a frictionless plane inclined at an angle $$\vartheta$$ above the horizontal, in the presence of a uniform gravitational field g. The block is realsed from rest at a distance L from the end of the plane. Use the energy method to derive an expression for the position x of the block on the plane as a function of time by evaluating t=$$\((dx/dt)^{-1}$$ dx

Homework Equations

The Attempt at a Solution

Okay so I took the integral of t along starting from L-->0. My answer was t(L)=-(dtL/dx) which can be simplified to t(L)= -(1L/v). Am I correct in my thinking that the integral of t would give the position of the block as a function of t.

 

[Redbelly98]

Homework Statement

A block slides on a frictionless plane inclined at an angle $$\vartheta$$ above the horizontal, in the presence of a uniform gravitational field g. The block is realsed from rest at a distance L from the end of the plane. Use the energy method to derive an expression for the position x of the block on the plane as a function of time by evaluating t=$$\((dx/dt)^{-1}$$ dx

Technically that should be

dt = (dx/dt)^-1dx​

Guess that was simply a typo on your part.

Homework Equations

For this problem, you'll need an equation that relates to the energy of the block.

The Attempt at a Solution

Okay so I took the integral of t along starting from L-->0. My answer was t(L)=-(dtL/dx) which can be simplified to t(L)= -(1L/v). Am I correct in my thinking that the integral of t would give the position of the block as a function of t.

The mistake here is in treating (dx/dt) as a constant when you do the integral. But it is not constant, the block is accelerating hence v changes.

You'll need to write (dx/dt) as a function of x, and then do the integral. See my comment after "Relevant equations" above.

 

[anubis01]

yeah it was supposed to be dt, my mistake.

okay well it took me a while (I had trouble with the integration) but I think I got the right answer, and I'll just upload an image instead because I'm not good with the P.F's latex formulas.

http://img21.imageshack.us/img21/4167/cci2709200900000.jpg

 

[Redbelly98]

You have the right idea, and are very close to the correct solution.

Can you account for the slope of the plane in the expression for U(x)?

 

[anubis01]

sorry for the rather late reply but it took me a while to figure out the equation. I know u(x)=mgh and I tried to approximate for h, I'm not to sure about my answer but here is my work redone.

http://img22.imageshack.us/img22/4167/cci2709200900000.jpg

 

[anubis01]

yeah my work above was wrong, I was just overcomplicating things. U=-mgsin$$\theta$$

here is my work redone

http://img297.imageshack.us/img297/6828/cci2809200900000.jpg

 

[Redbelly98]

Hi,

You've made some more progress but there is still a problem with the U(x) expression.

You had come up with

U = mgh​

which is correct.

In your latest post, you also wrote

U = mg sinθ​

implying that h=sinθ, which is not correct. Also, the units do not work out: U has units of energy, while mg·sinθ has units of force. The two sides of any physics equation must have consistent units; if they do not then something is definitely wrong.

So ... what is h, in terms of x and θ? That's what you need to work out, using some trigonometry. Then substitute that expression for h into

U = mgh​

Hopefully then it will work out.

 

[anubis01]

okay i got h=Lsin$$\theta$$ (I just replaced the x with L giving the wording of the question. So then U=mgLsin$$\theta$$ and I redid the work.

http://img198.imageshack.us/img198/6828/cci2809200900000.jpg