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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 headon (as in the figure).
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 headon (as in the figure).

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inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444253632076:dw

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2What 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.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2I did the system 1/2 mv^2=1/2kx^2+u m g x for energy conservation

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2u m g=k x for spring remaining compressed at the maximum compressed position

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Cool 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Well 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.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2the 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?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0So wait, is u your coefficient of static friction?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2(sorry for delayed responses, I'm working on other parts of this assignment that aren't so conceptual and more tedious)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2well I got du/dx=(kx+umg)/(mgx)

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2which, as expected, gives me kx=umg hahaha

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2then 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))

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2oh wait that's stupid, that assumes u_s=u_k which is totally not true I'm stoopid

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2yeah I got confused cause I assumed u_s=u_k for some reason and then subsituted that equation into the energy equation.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2figured it out now tho

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Cool question though, thanks for sharing :)
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