How Long Does a Box Slide Up a Ramp?

  • Thread starter Thread starter Westin
  • Start date Start date
  • Tags Tags
    Box Ramp
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 3K views
Westin
Messages
87
Reaction score
0

Homework Statement



You are helping your friend move, using a ramp to move boxes from the ground to the moving truck. You give a 26kg box a shove so it moves at 1.4m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs at the bottom of the ramp. The angle that the ramp makes with the ground is 31[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/100/char0E.png. The coefficients of static and sliding friction are 0.39 and 0.1, respectively.

a) For how much time does the box slide up the ramp?

[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmr10/alpha/144/char01.pngt=


b) How far up the ramp does the box slide?

[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmr10/alpha/144/char01.pngx= c) When the box stops sliding up the ramp, does it remain stopped or begin sliding back down?

d) Explain how you determined your answer for part (c):
[/B]

Homework Equations



u = a/g
a = delta v / delta t
other friction equations
[/B]

The Attempt at a Solution



.39*9.8*26 = 99.372
.1*9.8 = 25.48

I believe you need to sin31 and set up an equation that compares the two frictional forces together to get an answer but I am having trouble setting it up.
 
Last edited by a moderator:
on Phys.org
You have to think what makes it to stop.
Add all the forces that bring it to halt.
Use SUVAT equations for constant force/acceleratipn.
 
You can also try another method to test your concept understanding.
Consider the initial energy imparted and equate to work done by friction ans gravity; and change in potential.
 
Second attempt: a = 9.81*sin(31)+.1*9.81*cos(31) = 5.89m/s^2

(Vi)^2 = 2as

s = ((1.4)^2)/(2*5.89) = .17m distance travled

Still giving me an incorrect answer..
 
Westin said:
Second attempt: a = 9.81*sin(31)+.1*9.81*cos(31) = 5.89m/s^2

(Vi)^2 = 2as

s = ((1.4)^2)/(2*5.89) = .17m distance travled

Still giving me an incorrect answer..
Looks right to me (.166 to 3 digits). Do you know what the answer is supposed to be?