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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0oh, 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?

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0hopefully :) I seem to come up with 2 different answers; and i know one of them is wrong.

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

amistre64
 2 years ago
Best 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 \[P_eK_e\]

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

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

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

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

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

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

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0just 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?

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

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

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

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

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

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

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0i 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?

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0thanks :) 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.

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

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

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