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.
Well, I don't know how to resolve the problem with μ(x)=αx.
If somebody could help with some advise.
first learn the spelling of 'PHYSICS"!!!
haha it's fun really. Do u a stand-up showan?
ok, physics.....now some advise???
please
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.
ok, thank you Damon.
but friction is a non-conservative force, so conservation of energy cannot be applied
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.
oh yes it can if you figure out how much is dissipated in frictional heat.
I need to confirm my answer please:
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.
please somebody
I don't have part b) but hum....where do you take in consideration the friction force?
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???
have a look here:
physicsforums com/showthread.php?s=7ff50c016352c4bdec599201ad9df2cf&p=4628168#post4628168
I'm stuck with time too
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
??????
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
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
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???
please
t_1 = pi / (2*w) where w = sqrt((k+alpha*m*g)/m)
But t is not..... t = 2*pi/w
Anonymous, did you confirm your answer?
And thanks for your help
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
Hawk,
c) The block will move back and reach a second stop somewhere between x=0 and x=x1. is WRONG
for part c the correct answer is the first option
for c) it depends on your value!
if your spring and friction force are the same the block stays put, otherwise no!
so, what is the correct answer for c) ???
Please
the answer for c was posted by KS
check on the my post here
h t t p://w w w
jiskha . com
/display.cgi?id=1389359012