How Do You Prove the Equation and Find the Block’s Speed at Position L?

  • Thread starter Thread starter Ruinmaniac
  • Start date Start date
  • Tags Tags
    Calculus Dynamics
AI Thread Summary
The discussion revolves around a physics problem involving a block pushed by a constant force across a surface with a variable coefficient of kinetic friction. The user seeks to prove the equation ax = vx dvx/dx and find the block's speed at position L. They express confusion about how to approach the proof and the subsequent calculation. Another participant suggests integrating the equation and relating it to conservation of energy to derive the speed. The user concludes with a proposed final answer for the block's speed, indicating a correction after a factoring error.
Ruinmaniac
Messages
5
Reaction score
0

Homework Statement



A block of mass m is at rest at the origin at t = 0. It is pushed
with constant force Fo from x = 0 to x = L across a horizontal
surface whose coefficient of kinetic friction is μk = μ0(1 - x/L).
That is, the coefficient of friction decreases from μ0 at x = 0 to
zero at x = L.

Prove that:

ax = vx dvx/dx

Then, find an expression for the block’s speed as it reaches position L.

Homework Equations



F = ma
Fk = μkn

The Attempt at a Solution



I actually have no idea how to even approach this question, it was in the challenge problem set of my class year 1 physics textbook.

I know that vx = dx/dt, and if I was to sub that into the equation I have to prove, it makes sense, but I do not know how to actually prove it. For the second part, I actually have no idea whatsoever. Thanks for helping the poor new engineering student.

-Question taken from Physics for Scientists and Engineers 3rd Ed, Knight-
 
Physics news on Phys.org
Ruinmaniac said:
I know that vx = dx/dt,
... and ax = dvx/dt.
What do you get if you combine the two equations so as to eliminate dt?

Note that if you integrate the target equation and multiply both sides by the mass you will get a conservation of energy equation. You can use this to answer the question.
 
Thanks, I think I got it. Sorry for the bad quality photos, my webcam is the only camera I have when doing late night physics :P

1.
https://www.physicsforums.com/attachment.php?attachmentid=71883&stc=1&d=1407216234
2.
https://www.physicsforums.com/attachment.php?attachmentid=71884&stc=1&d=1407216234
3.
https://www.physicsforums.com/attachment.php?attachmentid=71885&stc=1&d=1407216234

Can you tell me if my final answer is right? There is no answer for this question in the textbook. The true final answer (after that factoring error I have) is:

vx = root(L(Fo/m - uog))
 

Attachments

  • picture751.jpg
    picture751.jpg
    55.5 KB · Views: 909
  • picture752.jpg
    picture752.jpg
    56.5 KB · Views: 912
  • picture753.jpg
    picture753.jpg
    50.9 KB · Views: 935
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top