Conservation of Energy and maximum height

Almoore01
Messages
12
Reaction score
0
Hi, here's another problem I'm a little stuck on:

A block of mass M slides up an incline with an initial speed Vi in the position shown. (The figure is of a block sliding up an incline at an angle "theta" with a final height of "H."

Part A: If the incline in fritionless, determine the maximum height H to which the block will rise, in terms of the given quantities and appropriate constants.

Part B: If the incline is rough with a coefficient of sliding friction "u" (mew), determine the maximum height to which the block will rise in terms of "H" and given quantities.

For part A I just used Conservation of Energy and got the maximum height "H" to be H = (((Vi)^2)/(2g)), which is the right answer.

For part B, I know I have to use Work = (Change in K + Change in U). So I have:

(umg(cos (theta))) * distance = ((1/2)((Vf)^2) - (1/2)((Vi)^2)) + (mgHf - mgHi). The masses canceled out and I assumed that the initial height and initial and final velocities equaled 0, so then my answer ended up being Hf = ucos(theta) *d, which isn't the right answer...

Any suggestions?
 
Last edited:
Physics news on Phys.org
Why did you make the initial speed zero?
 
When I tried to work the problem through the first time, I couldn't get the Vi to reduce to anything or cancel out, so then I tried to go back through with the assumption that Vi = 0 because there's no Vi in the final answer.
 
It seems that Vi is a given in the problem. How do you reason that there is no Vi in the final answer?
 
I have that the final answer is h = H/(1 + ((u)/tan(theta))) (given by teacher.) I'm having trouble getting to that answer. If I don't assume that Vi is zero, I have (using Work done): (u = coefficient of sliding friction.)

W = delta K + delta U. (umg (cos(theta)) * d) = ((1/2)m((Vf)^2) - (1/2)m((Vi)^2)) + (mgHf - mgHi). Mass cancels, therefore: ug(cos(theta)) * d = ((1/2)((Vf)^2) - (1/2)((Vi)^2)) + (gHf - gHi). Initial height and final velocity are 0 (?) so then: ug(cos(theta)) * d = -(1/2)((Vi)^2) + gHf...

And from there...I'm stuck. I don't know how to make "d" and "Vi" cancel.
 
What is h?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top