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.