# Box coming to a stop on a ramp

1. Feb 13, 2015

### Westin

1. The problem statement, all variables and given/known data

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?

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

c) When the box stops sliding up the ramp, does it remain stopped or begin sliding back down?

2. Relevant equations

u = a/g
a = delta v / delta t
other friction equations

3. 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: May 7, 2017
2. Feb 13, 2015

### Simon Bridge

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. Feb 14, 2015

### azizlwl

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. Feb 14, 2015

### Suraj M

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. Feb 15, 2015

### Westin

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. Feb 15, 2015

### haruspex

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