amistre64
  • amistre64
Object sliding down a ramp .... PE and KE
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
|dw:1348751150306:dw|
amistre64
  • amistre64
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?
amistre64
  • amistre64
this problem involves friction

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I think we can use work-kinetic energy theorem.
amistre64
  • amistre64
hopefully :) I seem to come up with 2 different answers; and i know one of them is wrong.
anonymous
  • anonymous
The aceleration must be constant right?
amistre64
  • amistre64
\[a=\frac{2d}{t^2}\] \[v=at=\frac{2d}{t}\] \[K_e=\frac12mv^2\] energy lost to friction \[P_e-K_e\]
amistre64
  • amistre64
assume a constant acceleration yes
anonymous
  • anonymous
You're did it correct i think
anonymous
  • anonymous
do you know the co-efficient of sliding friction?
amistre64
  • amistre64
not outright; but that can be calculated with given information i believe
anonymous
  • anonymous
Hmm, actually we don't need \(\mu\) since time is known.
anonymous
  • anonymous
mmm true.
amistre64
  • amistre64
\[\mu_k=\frac{g~sin\theta-a}{g~cos\theta}\] \[F_k=\mu_k*\vec n=\frac{g~sin\theta-a}{g~cos\theta}*mg~cos\theta\] \[F_k=m(g~sin\theta-a)\] \[W_f=F_k*d=md~(g~sin\theta-a)\]
anonymous
  • anonymous
hey you know that v=2d/t^2. So KE is just 1/2 mv^2.
anonymous
  • anonymous
i mean, what else do you want?
amistre64
  • amistre64
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?
anonymous
  • anonymous
The problem maybe 1.638 sec isn't the correct time :)
anonymous
  • anonymous
yes, v=3m/s in that case.
amistre64
  • amistre64
its pretty good, we slid the penny 5 times and averaged it out
amistre64
  • amistre64
at first the velocity of the Ke i just worked out threw me for a loop
anonymous
  • anonymous
you actually measured 3m/s!?
amistre64
  • amistre64
i couldnt see why the velocity from v=2d/t wasnt matching up :)
amistre64
  • amistre64
the 3 m/s was not measured, it was calculated
anonymous
  • anonymous
oh i thought you performed an experiment with a penny..
amistre64
  • amistre64
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?
anonymous
  • anonymous
yes. This works no matter what the friction is (even 0).
amistre64
  • amistre64
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.
amistre64
  • amistre64
..and a change in time, but thats a given :)
anonymous
  • anonymous
yes.
anonymous
  • anonymous
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
amistre64
  • amistre64
lol, great; i minute ten watching a sock slide down a banister ....
anonymous
  • anonymous
x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.