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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348751150306:dw
 2 years ago

amistre64 Group TitleBest 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?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
this problem involves friction
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.0
I think we can use workkinetic energy theorem.
 2 years ago

amistre64 Group TitleBest 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.
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.0
The aceleration must be constant right?
 2 years ago

amistre64 Group TitleBest 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 \[P_eK_e\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
assume a constant acceleration yes
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.0
You're did it correct i think
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
do you know the coefficient of sliding friction?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
not outright; but that can be calculated with given information i believe
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.0
Hmm, actually we don't need \(\mu\) since time is known.
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
mmm true.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[\mu_k=\frac{g~sin\thetaa}{g~cos\theta}\] \[F_k=\mu_k*\vec n=\frac{g~sin\thetaa}{g~cos\theta}*mg~cos\theta\] \[F_k=m(g~sin\thetaa)\] \[W_f=F_k*d=md~(g~sin\thetaa)\]
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
hey you know that v=2d/t^2. So KE is just 1/2 mv^2.
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
i mean, what else do you want?
 2 years ago

amistre64 Group TitleBest 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?
 2 years ago

imron07 Group TitleBest ResponseYou've already chosen the best response.0
The problem maybe 1.638 sec isn't the correct time :)
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
yes, v=3m/s in that case.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
its pretty good, we slid the penny 5 times and averaged it out
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
at first the velocity of the Ke i just worked out threw me for a loop
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
you actually measured 3m/s!?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i couldnt see why the velocity from v=2d/t wasnt matching up :)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
the 3 m/s was not measured, it was calculated
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
oh i thought you performed an experiment with a penny..
 2 years ago

amistre64 Group TitleBest 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?
 2 years ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.2
yes. This works no matter what the friction is (even 0).
 2 years ago

amistre64 Group TitleBest 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.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
..and a change in time, but thats a given :)
 2 years ago

rajathsbhat Group TitleBest 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
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
lol, great; i minute ten watching a sock slide down a banister ....
 2 years ago
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