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

- amistre64

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

- katieb

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- amistre64

|dw:1348751150306:dw|

- 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

this problem involves friction

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## More answers

- anonymous

I think we can use work-kinetic energy theorem.

- amistre64

hopefully :)
I seem to come up with 2 different answers; and i know one of them is wrong.

- anonymous

The aceleration must be constant right?

- amistre64

\[a=\frac{2d}{t^2}\]
\[v=at=\frac{2d}{t}\]
\[K_e=\frac12mv^2\]
energy lost to friction
\[P_e-K_e\]

- amistre64

assume a constant acceleration yes

- anonymous

You're did it correct i think

- anonymous

do you know the co-efficient of sliding friction?

- amistre64

not outright; but that can be calculated with given information i believe

- anonymous

Hmm, actually we don't need \(\mu\) since time is known.

- anonymous

mmm true.

- 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

hey you know that v=2d/t^2.
So KE is just 1/2 mv^2.

- anonymous

i mean, what else do you want?

- 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

The problem maybe 1.638 sec isn't the correct time :)

- anonymous

yes, v=3m/s in that case.

- amistre64

its pretty good, we slid the penny 5 times and averaged it out

- amistre64

at first the velocity of the Ke i just worked out threw me for a loop

- anonymous

you actually measured 3m/s!?

- amistre64

i couldnt see why the velocity from v=2d/t wasnt matching up :)

- amistre64

the 3 m/s was not measured, it was calculated

- anonymous

oh i thought you performed an experiment with a penny..

- 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

yes. This works no matter what the friction is (even 0).

- 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

..and a change in time, but thats a given :)

- anonymous

yes.

- 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

lol, great; i minute ten watching a sock slide down a banister ....

- anonymous

x)

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