Distance traveled sliding down incline

In summary, the conversation discusses a problem involving a box sliding down an incline with a given initial velocity and coefficient of kinetic friction. The work-energy theorem is used to solve for the distance the box slides before coming to rest, with the correct answer being 0.775 m. There was some confusion initially about using the work done by the parallel component of gravity, but the correct approach was eventually determined.
  • #1
learnitall
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?
 
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  • #2
learnitall said:
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
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
I forgot to mention that I'm using g=9.8 m/s^2 for gravity
 
  • #5
learnitall said:
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
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
learnitall said:
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
this makes a lot of sense now, thanks for the input!
 

What is the formula for calculating distance traveled when sliding down an incline?

The formula for calculating distance traveled when sliding down an incline is d = (mgh)/(mgsinθ), where d is the distance traveled, m is the mass of the object, g is the acceleration due to gravity, h is the height of the incline, and θ is the angle of the incline.

How does the angle of the incline affect the distance traveled?

The angle of the incline affects the distance traveled by changing the component of gravity that acts parallel to the incline. As the angle increases, the component of gravity acting parallel to the incline decreases, resulting in a shorter distance traveled.

Is the mass of the object a factor in the distance traveled?

Yes, the mass of the object is a factor in the distance traveled when sliding down an incline. The greater the mass, the greater the force of gravity and the longer the distance traveled.

What is the relationship between height and distance traveled when sliding down an incline?

The relationship between height and distance traveled when sliding down an incline is directly proportional. This means that as the height of the incline increases, the distance traveled also increases.

How does friction affect the distance traveled when sliding down an incline?

Friction affects the distance traveled when sliding down an incline by reducing the speed of the object. The greater the friction, the shorter the distance traveled as the object loses more energy due to friction. This can be accounted for by multiplying the distance traveled by a friction factor.

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