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Newton's laws and incline surfaces

  • Thread starter naada
  • Start date
1. The problem statement, all variables and given/known data

A box is given an initial velocity of 5m/s up a smooth 20 incline surface . The distance the box travel before coming to rest is?


2. Relevant equations

I can't solve it correctly , I can't get the idea of this question

3. The attempt at a solution

x= ?
vi=5
v=0
a=?

F = ma
mg cos ? = a

tan-1 20 = 87.14
180 = 90 + 87.14 + ?
?= 180 -90-87.14
?=2.9

(mg cos 2.9 = ma ) /m
g cos 2.9 = a
a= 28.03

x = 25 /( 28.03 )2
x= 0.45 m

My answer is so wrong !
But i can't figure out how to solve this problem
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

Doc Al

Mentor
44,684
1,003
A box is given an initial velocity of 5m/s up a smooth 20 incline surface .
Presumably a 20 degree incline?


F = ma
mg cos ? = a
What's the component of the weight acting down the incline?

tan-1 20 = 87.14
180 = 90 + 87.14 + ?
?= 180 -90-87.14
?=2.9
Not sure what you're doing here. I presume that the angle is given as 20 degrees.

(mg cos 2.9 = ma ) /m
g cos 2.9 = a
a= 28.03
You have the wrong angle (and trig function).


x = 25 /( 28.03 )2
x= 0.45 m
Sanity check: The acceleration due to gravity is only 9.8 m/s^2, so how can the acceleration parallel to the incline be greater than that?

Find the correct acceleration and that kinematic equation will work.

Read this to learn how to handle inclined planes: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
 
Last edited by a moderator:
Presumably a 20 degree incline?

the question didn't mention the word "degree" or " ْ "
it's exactly as I wrote it .
So I toke 20 as the slop

#
x=? vi=5 vf=0 a=?

F = ma
(m g cos20 = m a) / m
9.8 cos20 = a
a=9.21

x=25/(2a)
x=25/(18.42)
x=1.37 m
Which is also wrong

#
180 = 90+20+?
180 -90-20=70ْ

F= m a
(m g cos 70 = m a)/m
9.8 cos 70 = a
a=3.35

x=25/(2a)
x=25/(2*3.35)
x=3.37 m
I think this one is the right one,but not sure.

thanks for the link was so useful
 

Doc Al

Mentor
44,684
1,003
Presumably a 20 degree incline?

the question didn't mention the word "degree" or " ْ "
it's exactly as I wrote it .
So I toke 20 as the slop
I would take it to mean a 20 degree angle to the horizontal.

#
x=? vi=5 vf=0 a=?

F = ma
(m g cos20 = m a) / m
9.8 cos20 = a
a=9.21

x=25/(2a)
x=25/(18.42)
x=1.37 m
Which is also wrong
It's wrong because you took cos20, which is the component perpendicular to the incline. You need the parallel component.


#
180 = 90+20+?
180 -90-20=70ْ

F= m a
(m g cos 70 = m a)/m
9.8 cos 70 = a
a=3.35

x=25/(2a)
x=25/(2*3.35)
x=3.37 m
I think this one is the right one,but not sure.
This time you used the correct acceleration (which is g sin20), but you made a small error in your final calculation--probably a typo. (Redo the very last step.)

Other than that minor problem, you got it right.
 

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