## amistre64 Group Title Object sliding down a ramp .... PE and KE one year ago one year ago

1. amistre64 Group Title

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2. amistre64 Group Title

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?

3. amistre64 Group Title

this problem involves friction

4. imron07 Group Title

I think we can use work-kinetic energy theorem.

5. amistre64 Group Title

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

6. imron07 Group Title

The aceleration must be constant right?

7. amistre64 Group Title

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

8. amistre64 Group Title

assume a constant acceleration yes

9. imron07 Group Title

You're did it correct i think

10. rajathsbhat Group Title

do you know the co-efficient of sliding friction?

11. amistre64 Group Title

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

12. imron07 Group Title

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

13. rajathsbhat Group Title

mmm true.

14. amistre64 Group Title

$\mu_k=\frac{g~sin\theta-a}{g~cos\theta}$ $F_k=\mu_k*\vec n=\frac{g~sin\theta-a}{g~cos\theta}*mg~cos\theta$ $F_k=m(g~sin\theta-a)$ $W_f=F_k*d=md~(g~sin\theta-a)$

15. rajathsbhat Group Title

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

16. rajathsbhat Group Title

i mean, what else do you want?

17. amistre64 Group Title

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?

18. imron07 Group Title

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

19. rajathsbhat Group Title

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

20. amistre64 Group Title

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

21. amistre64 Group Title

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

22. rajathsbhat Group Title

you actually measured 3m/s!?

23. amistre64 Group Title

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

24. amistre64 Group Title

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

25. rajathsbhat Group Title

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

26. amistre64 Group Title

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?

27. rajathsbhat Group Title

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

28. amistre64 Group Title

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.

29. amistre64 Group Title

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

30. rajathsbhat Group Title

yes.

31. rajathsbhat Group Title

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

32. amistre64 Group Title

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

33. rajathsbhat Group Title

x)