Easy Acceleration / Friction question

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SUMMARY

The discussion centers on calculating the acceleration of a 50 kg box on a roof inclined at 10 degrees, with a coefficient of friction (µ) of 0.15. The user dglenn9000 initially calculates the gravitational force (Fg = 490 N), normal force (Fn = 482.56 N), and acceleration without friction (1.70 m/s²). The community advises applying Newton's second law to incorporate friction, specifically by calculating the frictional force (Ffriction = µ * Fn) and adjusting the total force equation (Ftotal = ma) accordingly to find the box's acceleration with friction.

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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
 
Last edited:
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Welcome to PF!

Hi dglenn9000! Welcome to PF! :wink:
dglenn9000 said:
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 :smile:
 
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?
 
Welcome to PF!

Hi DIrtyPio! Welcome to PF! :smile:

(but please start a new thread in future)
DIrtyPio said:
… 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. :wink:
 


tiny-tim said:
Hi dglenn9000! Welcome to PF! :wink:


Hint: just use good ol' Newton's second law …

total force (in the direction of the slope) = mass times acceleration :smile:


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?
 
Hi dglenn9000! :smile:

just woke up :zzz: …
dglenn9000 said:
… 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 :wink:
 
Hi everyone do you do any problems considering phyisc?
 

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