## inkyvoyd one year ago A m kg block slides along a horizontal surface with friction. The block has a speed v when it strikes a massless spring head-on (as in the figure).

1. inkyvoyd

|dw:1444253632076:dw|

2. inkyvoyd

What minimum value of the coefficient of static friction, µs, will assure that the spring remains compressed at the maximum compressed position? Assume the spring has a spring constant k.

3. inkyvoyd

I did the system 1/2 mv^2=1/2kx^2+u m g x for energy conservation

4. inkyvoyd

u m g=k x for spring remaining compressed at the maximum compressed position

5. Empty

Cool this looks fun, but the thing is we can't use energy conservation here because friction isn't a conservative force, so it's path dependent and not a state function.

6. inkyvoyd

fudge

7. Empty

Well let's try to piece this together, it looks like we can lay out a few variables and stuff and see what happens. Give me a sec to fiddle with some symbols on my paper here.

8. inkyvoyd

the first part of the problem specified the static friction, and asked what the compression would be, and I got that right using $$\large 1/2 mv^2=\int^x_0kx+umg \text{ }ds$$ which is how I got $$1/2mv^2=1/2kx^2+umgx$$ then I solved for x and put that in, and it said it was right... did I get it right cause I was lucky even though the method is invalid, or did that integral exhibit path dependency?

9. Empty

Yeah this is fine cause you are only going to the maximum, now if it were to go past that and say go back to spring equilibrium, then you'd feel the friction force again on the way back. So this integral is valid yeah.

10. Empty

So wait, is u your coefficient of static friction?

11. inkyvoyd

yeah

12. Empty

I think we might be able to take the derivative of u with respect to x and set it equal to 0 to find the minimum of u?

13. inkyvoyd

(sorry for delayed responses, I'm working on other parts of this assignment that aren't so conceptual and more tedious)

14. Empty

Yeah sure I'm just kinda having fun here anyways I haven't really touched this sorta stuff in a while. I'm not sure about this derivative though, I guess because v and x will be a constant at the maximum.

15. inkyvoyd

well I got du/dx=-(kx+umg)/(mgx)

16. inkyvoyd

which, as expected, gives me kx=umg hahaha

17. inkyvoyd

then I did the system; 1/2 mv^2=1/2kx^2+u m g x kx=umg and got u=(v sqrt(k))/(g sqrt(3m))

18. inkyvoyd

which... is wrong

19. inkyvoyd

oh wait that's stupid, that assumes u_s=u_k which is totally not true I'm stoopid

20. Astrophysics

That should work

21. Astrophysics

$kx= u mg$

22. inkyvoyd

yeah I got confused cause I assumed u_s=u_k for some reason and then subsituted that equation into the energy equation.

23. Astrophysics

Ah ok I see

24. inkyvoyd

figured it out now tho

25. inkyvoyd

thanks

26. Astrophysics

Cool question though, thanks for sharing :)