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Distance traveled sliding down incline

  • Thread starter learnitall
  • Start date
  • #1
10
2

Homework Statement


A box is sliding down an incline tilted at a 11.1° angle above the horizontal. The box is initially sliding down the incline at a speed of 1.70 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline before coming to rest.

a) 1.08 m
b)0.775 m
c)0.620 m
d)0.929 m
e)The box does not stop. It accelerates down the plane

Homework Equations


Work-Energy Therorem:
U1 + K1 + Wother = U2 + K2

K=(1/2)mv2
U=mgh

Work=Fdr
Ffriction=μFnormal


The Attempt at a Solution


Given an initial velocity and a coefficient of kinetic friction, immediately I thought of using the work energy theorem to solve this problem.

I first set up a coordinate system that is tilted and running along the plane, with the origin at the top of plane.

I assumed the block starts from the top (with given initial velocity) and ends at a distance d - the unknown variable we're solving for.

I drew a free body diagram of the box and summed the forces in both the X an Y directions.
Here is the result of that:

ƩFx= mgsinθ - Ffriction
ƩFy=Fnormal-mgcosθ=0


I ignored the Fy equation and used the Fx one to get:
Fnormal=mgcosθ

so now I set up my Work-Energy Theorem Equation exactly like how I wrote it in part 2 above:
0 + (1/2)mv2 - μmgcosθd = 0 + 0
The equation is set to zero because the block stops at the end of the interval (∴ K2= 0) and there is never potential energy as per my coordinate system (∴U2=0)

I solved my Work-Energy equation for 'd' and got 0.3852819047≈0.385.
Can anyone tell me where I went wrong? Did I take the wrong approach? which one should I take?
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
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and there is never potential energy as per my coordinate system
That's your problem. Just because you chose a coordinate system parallel to the incline does not mean that potential energy is always zero. Gravity still acts vertically.
 
  • #3
10
2
Ok so if I consider the work done by gravity's horizontal component through the distance 'd' as -mgdcos(theta) and plug that value into the left side of my Work-Energy equation.
My new result is:
d = .1087006786 m or approximately .108 m.
My new result is off by a factor of 10 from the choice 'a) 1.08 m'
Was I right by using the work done by the parallel component of gravity? I didn't use the vertical component because I know its perpendicular to the distance traveled.
 
  • #4
10
2
I forgot to mention that I'm using g=9.8 m/s^2 for gravity
 
  • #5
Doc Al
Mentor
44,905
1,169
Ok so if I consider the work done by gravity's horizontal component through the distance 'd' as -mgdcos(theta) and plug that value into the left side of my Work-Energy equation.
You are correct in using the component of gravity parallel to the incline in finding the work done by gravity. But the parallel component of gravity is not mgcosθ. See your first post!

Correct that mistake and you'll be fine.
 
  • #6
10
2
You were right. I was missing the work don't by the parallel component of gravity which was +mgsin(theta)d. Using this quantity for the equation gives : d= .775 which is answer 'b'.
Thank you Doc Al
 
  • #7
3,740
417
Work-Energy Therorem:
U1 + K1 + Wother = U2 + K2
Maybe you will avoid confusion in the future by using the usual format of the Work-Energy theorem.
What you wrote is not the work-energy theorem, even if mathematically is equivalent.
The WE theorem state that the change in KE (only kinetic) equals the work done by all the forces acting on the body. So you don't have to worry about which forces are conservative and potential energy.
 
  • #8
10
2
this makes alot of sense now, thanks for the input!
 

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