Introduction to How Path Independence Leads to Potential Energy
Follow-up to Problem 44 from the Book as handled by previous notes:
From page 153, #44. Read it, or do the previous notes:
kid starts at this height
reference line
· Point A will be called the starting point on the slide. Its height is D above some reference line.
· Point B will be called the lowest point on the slide. Its height is H above the same reference line.
· Point C will be called the place where the girl is launched off of the slide. The altitude of point C is d above the same reference line.
· As is true in the book’s diagram, it is given that (d – H) = h/5. Please confirm that I did this right. Also, D – H = h is a given. Also, h = 4 meters is a given.
· Please label all of these locations and heights in the diagram above.
· Notice how randomly the reference line is chosen. THAT IS EXTREMELY IMPORTANT. The book’s problem lacks this randomness, so these notes are better.
I’ve made the diagram more sophisticated, so use the one on this page.
Focus now on the problem of finding the girls launch speed at point C. Even if you got this correct before, you are to do these new notes pretending you haven’t done it before (but know how to set it up.)
vC = ?
By following my advice in class, you should now have several solutions to this, all with careful set-up using the work energy theorem. Each solution method practices with different definitions of initial and final point. (One method is not enough when testing for understanding.)
On the following pages, make sure you know how to intelligently fill in a proper sign and a proper position change expression. I’ve left blanks for you to do this. That is why there is a blank left in front of expressions. That is for the clearly written sign. Also, for the K side of things, know what you may assume and what you must leave as undetermined.
All set-up expressions are from the Work-Energy Theorem.
B to C set up: WB→C = KC – KB
( )Mg( ) = KC – KB
( )Mg( ) = ( ) – ( )
I have no way of knowing what KC and KB are. Neither motion at those instants is at rest, so I can’t claim to know what to plug in. So I just leave them called KB and KC. Still, the B→C setup is important to write.
A to C set up: WA→C = KC – KA
( )Mg( ) = KC – KA
( )Mg( ) = ( ) – ( )
Even if confused on how to start the problem, something in the clear set up above ends up leading to the conclusion that KC = Mg(D – d) if all of your subs were accurate. That’s pretty useful, because we know M, g, D, and y. Therefore, we know KC, and since KC=½MvC2, we now know vC. Problem solved. Pretty neat solution, eh?
Make sure that your subs into some of the blanks above confirm what I just said about how to solve for vC. Make sure you know how to do it. It leads to the launch speed we got before, 8 m/s.
Ok, that’s fine, but I want you to see things a different way. Set up A to B. That hasn’t been done yet, unless you did it yourself, as you should have.
A to B set up: WA→B = KB – KA
( )Mg( ) = KB – KA
( )Mg( ) = ( ) – ( )
This is very important: Pretend that you had set up B to C, and it got you nowhere (as it did above.) Now look at the set-up from B to C, and then this set-up from A to B, and before I spell it out for you, write what is important about this comparison. Write it below for yourself. Only read the next page after you have figured out the importance for yourself. I give it away on page 3.
(By the way, it is required that your writing habits in solutions are as clear and thorough as on this page.)
Precise Solution Expressions:
B to C set up: WB→C = KC – KB
(-)Mg(d – H) = KC – KB
(-)Mg(d – H) = KC – KB
A to B set up: WA→B = KB – KA
(+)Mg(D – H) = KB – KA
(+)Mg(D – H) = ( KB ) – ( 0 )
Here’s the deal. Take that expression you see for KB right there and go plug it into the “KB” term that lies within the A to B set-up. Then solve for KC, simplify, and see what happens. Why is this significant?
It ends up communicating the exact same thing that the direct A to C set-up communicates. This should be highlighting the importance of path-independence for you.
A to B set up: WA→C = KC – KA
( + )Mg(D – d) = KC – 0
Stay tuned for the next part: how this leads to the goodness of defining Potential Energy.