# physics (Urgent)!!!!!!!!!!!!!!!!!!!!!!

A small block of mass m=1 kg glides down (without friction) a circular track of radius R=2 m, starting from rest at height R. At the bottom of the track it hits a massless relaxed spring with spring constant k= 9 N/m, which starts to be compressed as the block continues to move horizontally. Note that we assume no energy loss during this “collision". There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 0.6 m−1. Assume for simplicity that static and dynamic friction coefficients are the same, and use g=10 m/s2
(a) What is the maximal distance x1 that the block moves horizontally away from the track at x=0? (in meters)

x1=
(b) What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds)

t1=
(c) What will happen after the block reaches point x1?

The block will move back and get catapulted up the circular track.

The block will move back and reach a second stop somewhere between x=0 and x=x1.

The block will move back and reach a second stop exactly at x=0.

The block will stay put forever at x=x1.

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1. Well, I don't know how to resolve the problem with μ(x)=αx.

If somebody could help with some advise.

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posted by NTEA
2. first learn the spelling of 'PHYSICS"!!!

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3. haha it's fun really. Do u a stand-up showan?

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posted by NTEA
5. I advise you to learn about conservation of energy and integrals of F dot dX and put some time into solving this 8:01 final question yourself.

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posted by Damon
6. ok, thank you Damon.

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posted by NTEA
7. but friction is a non-conservative force, so conservation of energy cannot be applied

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8. I've spent 4 hours to solve it as the result question a is done, but i'm a russian teenager, that is going to attend the uneversity, and unfotunately, i don't know how to solve it with non-constant friction koff.

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posted by Anon
9. oh yes it can if you figure out how much is dissipated in frictional heat.

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posted by Damon
For a:
First for the theorem W_E:
The work is for the friction and for the spring (left side of the eq.) and the energy only for the potential(right side)
I obtained:
-ux^2/2 - kx^2/2 = -mgh
then the x=1,....

For b:
I analized the part x=0 to x=x1
mV^2/2 = kx^2/2
Integrating.....
t= 0.0......

for c:

The block will move back and reach a second stop somewhere between x=0 and x=x1.

Is that correct??? is my last op. please.

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posted by Hawk

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posted by Hawk
12. I don't have part b) but hum....where do you take in consideration the friction force?

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13. for part b, I considered only the conservation of energy.I think that the friction is only for work-energy analysis.
But I am not so sure???

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posted by Hawk
14. have a look here:

I'm stuck with time too

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posted by some0ne
15. I did it in this way:
using the theorem W-E
for the point x=0,x=x1:

kx^2 /2 - mV^2 /2 = -αmgx^2 /2

algebra,resolving for V:

V= x*sqrt((αmg + k)/m)

We know that V = dx/dt

so:

dx/x = (sqrt((αmg + k)/m))dt

ln x = (sqrt((αmg + k)/m))t

reolving for t:

t= ln (1.085)/sqrt(34)

t=0.014 s

but it is a little expression

??????

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posted by Hawk
16. I used the W-E theorem too,basically:
-> Potential Energy became Kinetic Energy at the end of the track
-> in the collision no energy is lost as it is stated by the problem, so
-> it is KE = Work F_spring + Work F_friction
-> they both change with x so i took the integral from x0 to x1 for both the Work done
that way I found max displacement.

But without using SHO I'm at a loss to find the time.

For c) start by seeing if , like for me, spring force and friction force are the same, if som the block stay put at x1

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posted by some0ne
17. sorry I told you wrong for c)...
if friction and spring are the same, given friction static and kinetic coefficient is the same, it will move back

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posted by some0ne
18. I don't know b.
For a) x=1.085
c) The block will move back and reach a second stop somewhere between x=0 and x=x1.

are those correct???

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posted by Hawk

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posted by Hawk
20. t_1 = pi / (2*w) where w = sqrt((k+alpha*m*g)/m)

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21. But t is not..... t = 2*pi/w

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posted by ans
22. Indeed, period formula is 2pi/w but how much of the period you need?
It depends on your data, if the block stay put at x1 you need only 1/4 of period, if the block comes back you need more

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posted by some0ne
23. Hawk,
c) The block will move back and reach a second stop somewhere between x=0 and x=x1. is WRONG

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posted by Greco
24. for part c the correct answer is the first option

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posted by KS
25. for c) it depends on your value!

if your spring and friction force are the same the block stays put, otherwise no!

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posted by some0ne
26. so, what is the correct answer for c) ???

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posted by ans
27. the answer for c was posted by KS

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posted by Hawk
28. check on the my post here

h t t p://w w w

jiskha . com

/display.cgi?id=1389359012

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