Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Object sliding down a ramp .... PE and KE

Physics
See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
|dw:1348751150306:dw|
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?
this problem involves friction

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question