a 1210kg rollercoaster car is moving 8.31 m/s. before entering the station, it first rolls up a 2.26m hill. how fast is it moving then? HEELLLPPPPP!!!
we know that kinetic energy = \(\frac{1}{2}mv^2\) and potential energy = \(mgh\) where \(g \approx 9.8 m/s^2\) Also, kinetic energy + potential energy is constant, ignoring frictional losses from the wheels, wind resistance, etc. We have no way of computing those things, so we'll assume they are to be ignored. when the car is moving at the start of the problem, the total energy is just its kinetic energy at that point. use the first formula to compute it. After rolling up the hill, some of the kinetic energy is converted to potential energy. Compute the potential energy at that point and subtract it from the original kinetic energy. The remaining energy is kinetic, and you can work backwards to find the velocity that corresponds to that amount of kinetic energy with the given mass. Try it and see what you get, I'll check your work if you want.
well i used the kinetic energy formula and then i did the potential energy and subtract them. when it comes to the velocity i dont know what to do
\[KE = \frac{1}{2}mv^2\]Solve that equation for \(v\), then plug in the numbers.
so it would be v= ke/2*m?
\[KE = \frac{1}{2}mv^2\]Multiply both sides by 2\[2KE = mv^2\]Divide both sides by \(m\)\[\frac{2KE}{m}=v^2\]Take square root of both sides\[\sqrt{\frac{2KE}{m}}=v\]
with the answer that i got when i subtracted KE and PE right?
yes. use the remaining KE
which by my figures is about 14979.9 kg m^2/s^2
yes that is what i got
good!
4.97595217019 this is what i got as my final answer
Yeah, about that :-)
okay thank u
how about when it has degrees?
not reasonable to put that many significant digits in the answer, however — we've only got 3 significant figures in our measurements of mass, speed, height...
how about when what has degrees?
yeah i got that answer right but i dont know how to solve this problem now because it has degrees
I don't see anything in degrees in this problem...
hold up
a 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3 degress. at the bottom she is moving 20.3 m/s . how much work did friction on the woman? 42.6 is not her starting height
|dw:1371843934540:dw| use trig to find the height
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