Easy Acceleration / Friction question

[dglenn9000]

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?

Homework Equations

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

 

[tiny-tim]

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

 

[DIrtyPio]

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]

Welcome to PF!

Hi DIrtyPio! Welcome to PF!

(but please start a new thread in future)

… 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.

 

[dglenn9000]

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]

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: F_friction = µN.

You then chuck the friction force into the "total force" in F_total = ma

 

[mg.radebe]

Hi everyone do you do any problems considering phyisc?