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amistre64Best ResponseYou've already chosen the best response.0
dw:1348751150306:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
oh, and time it takes to slide to the bottom of the ramp is 1.638 seconds Potential energy = .0025*9.8*.46 Kinetic energy at the bottom of the ramp is?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
this problem involves friction
 one year ago

imron07Best ResponseYou've already chosen the best response.0
I think we can use workkinetic energy theorem.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
hopefully :) I seem to come up with 2 different answers; and i know one of them is wrong.
 one year ago

imron07Best ResponseYou've already chosen the best response.0
The aceleration must be constant right?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[a=\frac{2d}{t^2}\] \[v=at=\frac{2d}{t}\] \[K_e=\frac12mv^2\] energy lost to friction \[P_eK_e\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
assume a constant acceleration yes
 one year ago

imron07Best ResponseYou've already chosen the best response.0
You're did it correct i think
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
do you know the coefficient of sliding friction?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
not outright; but that can be calculated with given information i believe
 one year ago

imron07Best ResponseYou've already chosen the best response.0
Hmm, actually we don't need \(\mu\) since time is known.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[\mu_k=\frac{g~sin\thetaa}{g~cos\theta}\] \[F_k=\mu_k*\vec n=\frac{g~sin\thetaa}{g~cos\theta}*mg~cos\theta\] \[F_k=m(g~sin\thetaa)\] \[W_f=F_k*d=md~(g~sin\thetaa)\]
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
hey you know that v=2d/t^2. So KE is just 1/2 mv^2.
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
i mean, what else do you want?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
just working it thru a couple of ways. When I first tried to work this out; I used: Pe = mgh = 0.01127 J in a frictionless system; Ke = Pe at the bottom of the ramp mgh = mv^2/2 gh = v^2/2 sqrt(2gh) = v v = 3.0027 m/s right?
 one year ago

imron07Best ResponseYou've already chosen the best response.0
The problem maybe 1.638 sec isn't the correct time :)
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
yes, v=3m/s in that case.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
its pretty good, we slid the penny 5 times and averaged it out
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
at first the velocity of the Ke i just worked out threw me for a loop
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
you actually measured 3m/s!?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i couldnt see why the velocity from v=2d/t wasnt matching up :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the 3 m/s was not measured, it was calculated
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
oh i thought you performed an experiment with a penny..
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i spose my question is; is using the velocity of 2d/t a proper way to determine the kinetic energy in the system at the bottom of the ramp?
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
yes. This works no matter what the friction is (even 0).
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
thanks :) so if it was a frictionless system, then the kinetic energy at the bottom of the ramp would still equal the potential energy from the start of it; and all that would change is the final velocity of the object then.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
..and a change in time, but thats a given :)
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.2
Here's a nice video of an extreme case where the object slides with uniform velocity: http://www.youtube.com/watch?v=84DXjumAxNQ&feature=plcp
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
lol, great; i minute ten watching a sock slide down a banister ....
 one year ago
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