How many times will the block cross the rough patch


by JNason11
Tags: energy, frictionless, kinetic energy, physics, work
JNason11
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#1
Nov19-12, 06:27 PM
P: 5
A 7.50kg block slides down a frictionless ramp of height 2.25m at a velocity of 1.50 m/s. it crosses a rough patch that is 2.5m long and exerts a force of 3N. It then continues to the other side of the ramp which is angled at 35 degrees(I dont think this matters b/c it is frictionless). How can I find how many times the block will cross the rough patch? I tried a few equations which didn't make sense....

W=ΔE=Ef-Ei
W=FdCosθ
KE=1/2mv^2
PE=mgh



I tried to find the work being done and used Fdcos180 because the patch acts against the direction of motion so it would be -Fd where F=3N and d=2.5m.
Then I set Ef-Ei=W and then it started to fall apart. I am just not sure how to approach this problem I guess... Ef would be 0 if we are looking for how many times it crosses the rough patch right? so then -Ei=-Fd, but I just am having trouble relating things... help would be appreciated, thank you!
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TSny
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#2
Nov19-12, 06:47 PM
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Quote Quote by JNason11 View Post
Ef would be 0 if we are looking for how many times it crosses the rough patch right? so then -Ei=-Fd
OK, but what does d represent here: the distance traveled in going once across the rough patch or the total distance traveled on the rough patch before coming to rest?
JNason11
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#3
Nov19-12, 06:59 PM
P: 5
going once across the rough patch

TSny
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#4
Nov19-12, 07:07 PM
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How many times will the block cross the rough patch


Does going once across the patch reduce the mechanical energy E all the way to zero? If not, then you cannot let Ef = 0 after one crossing.
JNason11
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#5
Nov19-12, 07:22 PM
P: 5
We are supposed to find the total work done and then find the work done by 1 crossing and divide total/1 crossing to find the number of crossings across the frictional surface. It will cross more than once, i am sure of this because everything else in the system is frictionless
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#6
Nov19-12, 07:32 PM
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The total work will bring the object to rest. The total work is determined by the total distance D that the object travels on the rough patch. How would you express the total work in terms of F and D? How would that total work relate to the initial energy?
JNason11
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#7
Nov19-12, 08:19 PM
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Here I attached the problem to make it easier. it is #3
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File Type: pdf HW Set 12 F12.pdf (40.0 KB, 5 views)
TSny
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#8
Nov19-12, 08:27 PM
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You are close to getting the answer. Let D be the total distance that the block travels across the rough patch. (It doesn't matter that this total distance might be broken up into several trips across the patch.) How would you write the total work done by the force F in terms of D and how would that work relate to the initial and final energies?

Once you solve for D, you can figure out how many times it crosses the patch and where it finally comes to rest.
JNason11
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#9
Nov19-12, 09:22 PM
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I guess I am just not getting it...
haruspex
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#10
Nov19-12, 10:12 PM
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What are the possibilities for where the block comes to rest?
How is its energy distributed between KE and PE in those areas?
In what ways does it lose total energy?
How rapidly (J/m) does it lose energy crossing the rough area?
TSny
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#11
Nov19-12, 10:23 PM
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You know that the total work done by the friction force F is W = -FD where D is the total distance traveled over the rough patch. (If the object traveled 5 and a half times over the patch before coming to rest, then D would equal 5.5d.) You also know that the object ends up with zero mechanical energy when it finally comes to rest somewhere on the rough patch. So, use your equation W = ΔE and see if you can find D. From that you can find the number of times it crosses the rough patch.


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