Box coming to a stop on a ramp

  • Thread starter Westin
  • Start date
  • #1
87
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 [Broken] 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. [Broken] 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= [Broken]


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

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


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:

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,654
You set up the equation from a free body diagram.
Note: the numbers you supplied are meaningless without some words to go with them... you also need units.
 
  • #3
1,065
10
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.
 
  • #4
Suraj M
Gold Member
599
39
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.
 
  • #5
87
0
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..
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,550
5,966
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?
 

Related Threads on Box coming to a stop on a ramp

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
2K
Replies
14
Views
11K
  • Last Post
Replies
14
Views
857
Top