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Kinetic and Potential energy problem

  • Thread starter BrainMan
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  • #1
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Homework Statement


A sled and its rider have a mass of 90 kg and a speed of 3 m/s at point A on the frictionless hill of Figure 5.25. At the bottom of the hill, the sled moves onto a rough surface with a coefficient of friction of .6. How far from the base of the hill will the sled move before stopping?


Homework Equations


KEf + PEf = KEi + PEi

V^2= Vo^2 - 2ax

KE= 1/2m(v^2)

PE= mgy

The Attempt at a Solution


What I attempted to do was to find the speed at the end of the ramp by using the formula
KEf + PEf = KEi + PEi
so
1/2(90)(9.8)((3)^2) + (90)(9.8)(3) = 1/2m(v^2)
v= 7.67

Then I used the velocity to find the total distance by using the formlula V^2= Vo^2 - 2ax
so
0 = 60.2176 - 2(54)(x)
// I got 54 by multiplying the coefficient of friction by the normal force//
when you solve for x you get .56 m. The real answer is 5.77 m

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
BiGyElLoWhAt
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Just use conservation of energy. It's way cleaner and easier. Find the total energy at the bottom of the hill just before the rider enters the rough surface, and use E0 - W = Ef = 0 (because the rider lost all potential energy in coming down the hill which was converted to kinetic energy, and then the rider and sled have -W done on them by the surface)
 
  • #3
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How do I find the distance the rider traveled then?
 
  • #4
BiGyElLoWhAt
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Work is ##\int_{0}^{d} F \cdot dx = \int_{0}^{d} F dx cos(\theta)##
Idk if you know how to do integrals or not, this is pretty simple. ##cos(\theta)## is a constant, and since friction acts opposite the velocity, theta is pi, and cosine theta is -1. F is a constant (##\mu N##) and ##\int_{0}^{d} dx = x|_{0}^{d} = d-0##

put it all together and work done by friction is ##\mu N d (-1) = -\mu N d##

Use that for W in conservation of energy, and solve for d.

Hope that helps. BTW what level physics/math are you in?
 
  • #5
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2
I do not understand integrals or calculus only algebra
 
  • #6
BiGyElLoWhAt
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that's ok, you know conservation of energy though right?
You know the kinetic energy at Point a, and you know the potential energy at point a. That means you know the total energy at point a. Since no work is done between a and just before the rider hits the rough patch, that means you know the total energy right before the rough patch, as energy is conserved.

I gave you the work formula for friction, so take the total energy just before the rough patch and plug it into the conservation of energy equation. Then take the formula for work that I gave you (which should have come up in class, I'm sure), and plug that in. Then set the whole thing equal to the total energy at the end.
*hint* What's the potential energy at the bottom of the hill, and what is the kinetic energy of the rider once he has stopped, which is what your looking for *un-hint*

As to the algebra, I figured as much, that's why I worked it out for you. =]
 
  • #7
BiGyElLoWhAt
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Ooo also, I noticed one slight flaw in your conservation of energy equation just now!

The proper equation to use is this:

##E_{total_{i}} \pm \Delta W = E_{total_{f}}##

Need that work in there bro.
 
  • #8
haruspex
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v= 7.67
Seems a bit low. Check your calculation.
0 = 60.2176 - 2(54)(x)
The mass is 90kg. What is the weight?
 
  • #9
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OK I figured out this problem in a different way. My formula for the conservation of energy is correct I looked it up. The reason I got the question wrong is one I calculated the final velocity wrong its actually 8.23 m/s two I didn't use the right formula for the deceleration due to friction. The way for calculating the acceleration due to friction when using a kinematic equation like I did is the use the formula a= -μg which is the acceleration equals the coefficient of friction times the force of gravity. I used it and found the correct answer.
 
  • #10
BiGyElLoWhAt
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That works, but I wouldn't rely too heavily on that energy equation. Think about it, is the total energy of the rider and sled the same as it was at point a? Definitely not. That equation only works if there's no work, but in this case there is, so it doesn't work for the friction part. Its good for from point a to the bottom of the hill, but after that its useless.
 
  • #11
haruspex
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That works, but I wouldn't rely too heavily on that energy equation.
I think you're confusing energy conservation (always true) with work conservation. There was nothing wrong with the method in the OP, just the details.
 
  • #12
ehild
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1. Homework Statement [/b]
A sled and its rider have a mass of 90 kg and a speed of 3 m/s at point A on the frictionless hill of Figure 5.25. At the bottom of the hill, the sled moves onto a rough surface with a coefficient of friction of .6. How far from the base of the hill will the sled move before stopping?


Homework Equations


KEf + PEf = KEi + PEi

V^2= Vo^2 - 2ax

KE= 1/2m(v^2)

PE= mgy

The Attempt at a Solution


What I attempted to do was to find the speed at the end of the ramp by using the formula
KEf + PEf = KEi + PEi
All good, but you need much less work if you solve the problem symbolically first.

so
1/2(90)(9.8)((3)^2) + (90)(9.8)(3) = 1/2m(v^2)
v= 7.67
write mgh+1/2 mvi2=1/2 mvf2.

You can multiply the equation by 2/m, and you get vf2=2gh+vi2=6*9.8+32=67.8, vf=8.23 m/s

Then I used the velocity to find the total distance by using the formlula V^2= Vo^2 - 2ax
so
0 = 60.2176 - 2(54)(x)
// I got 54 by multiplying the coefficient of friction by the normal force//
when you solve for x you get .56 m. The real answer is 5.77 m
"54" is not acceleration but force. To get acceleration, you divide the force of friction by the mass. Its magnitude is μg. Vo2is the same as vf2=67.8.

ehild
 

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