Recent content by jimithing71

LOVE YOU Doc Al! Thanks for all your help! Time for a glass of vino! :0)

Is that it?

I wish I knew! Okay, last attempt: KE(a) + PE(a) = KE(b) + PE(b) 0 + 176,400 = KE(b) + 105,840 + 30,000 176,400 - 105,840 - 30,000 = KE(b) 40,560 = KE(b) ?

It would be slowed due to friction? I am really lost on this one. I did the PE=mgh for point A and for point B and subtracted the two and got 70,560 so obviously that is wrong as well.

Sorry but I don't get that at ALL! That would be PE=(900kg)(9.8g)(20m), PE= 176,400? What am I missing?

Which one?

Or wait! Is it PE=mgh??

G=6.67X10^-11* N*m^2/kg^2 ??

Thank you! PE=1/2kx^6, 30=1/2k(.04), 30/.04=1/2k(.04)/.04, 750=1/2k, 1500=k Whew! One more? A roller coaster car with a mass of 900kg starts at rest from a point 20m above ground. At point B it is 12m above ground. If the initial KE was 0 and the work done against friction between the...

My head is starting to hurt! LOL! Okay, so now I tried F=-kx, 30J=-k(.20m), 30J/.20m=-k, 150=k That is one of the options (C), but our instructor told us the answer was (D) 1500. ?

To stretch a spring a distance of 0.20m 30J of work is done. What is the value of the spring constant k of the spring? a)6 b)30 c)150 d)1500 e)none of the above I know that the answer is D 1500 but I can't quite arrive at that. The two formulas I thought to try were F=-kx and...