# Easy Acceleration / Friction question

## Homework Statement

A roof has an incline of 10 degrees. If the µ of friction between a 50kg box and roof is .15, how fast will the box move?

Ff=µ * Fn

## The Attempt at a Solution

I have no idea how to get this. I know it very easy.
I figured out Fg = 490N (Fg=9.8 * 50kg)
I figured out Fn = 482.56 (Fn=490 cos 10)
I figured out F//=85.09 (F//=490 sin 10)

I figured out the Acceleration with no friction is 1.70 m/s2 85.09/50

Any help is appreciated, thanks

Last edited:

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tiny-tim
Homework Helper
Welcome to PF!

Hi dglenn9000! Welcome to PF! A roof has an incline of 10 degrees. If the µ of friction between a 50kg box and roof is .15, how fast will the box move?
Hint: just use good ol' Newton's second law …

total force (in the direction of the slope) = mass times acceleration I have a questions : there is an air track and on that track there is a body at rest upon which acts a constant force for a given time it gains a velocity. My question is if that force acts again upon that body for that amount of time will the velocity be 2 times the first velocity? My opinion is that starting from Newtons law v=v0+at, but if we use two times this quation we obtain that v1=at+at=2v0. Am I right?

tiny-tim
Homework Helper
Welcome to PF!

Hi DIrtyPio! Welcome to PF! … a constant force for a given time … if that force acts again upon that body for that amount of time will the velocity be 2 times the first velocity? My opinion is that starting from Newtons law v=v0+at, but if we use two times this quation we obtain that v1=at+at=2v0. Am I right?
yes, but it's better to look at it from the words of Newton's second law …

force = rate of change of momentum = mass times rate of change of velocity,

so (from the meaning of "rate of change") ∫force times time = mass times change of velocity. Hi dglenn9000! Welcome to PF! Hint: just use good ol' Newton's second law …

total force (in the direction of the slope) = mass times acceleration That is how I got how fast it will move without friction.

a= F/m
1.70 m/s2= 85.09/50kg

But my question is how do I figure out the acceleration with friction of .15? Where do I apply the .15 in my equation?

tiny-tim
Homework Helper
Hi dglenn9000! just woke up :zzz: …
… Where do I apply the .15 in my equation?
You multiply it by the normal force to get the (magnitude of the) friction force: Ffriction = µN.

You then chuck the friction force into the "total force" in Ftotal = ma Hi everyone do you do any problems considering phyisc?