# Box coming to a stop on a ramp

## 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?

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?

[/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.

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Simon Bridge
Homework Helper
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.

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.

Suraj M
Gold Member
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..

haruspex
Homework Helper
Gold Member
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?