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A 1210 Kg Roller Coaster

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Mathematics

OpenStudy (bearding):

a 1210kg rollercoaster car is moving viii.31 thou/s. before inbound the station, it first rolls upwardly a two.26m hill. how fast is it moving then? HEELLLPPPPP!!!

OpenStudy (whpalmer4):

we know that kinetic energy = \(\frac{i}{two}mv^ii\) and potential energy = \(mgh\) where \(thou \approx 9.8 m/s^2\) Also, kinetic energy + potential free energy is abiding, ignoring frictional losses from the wheels, wind resistance, etc. Nosotros accept no way of computing those things, and then nosotros'll assume they are to be ignored. when the car is moving at the start of the problem, the total energy is but its kinetic free energy at that betoken. use the showtime formula to compute it. After rolling upward the hill, some of the kinetic energy is converted to potential free energy. Compute the potential free energy at that signal and subtract it from the original kinetic energy. The remaining energy is kinetic, and y'all tin can work backwards to observe the velocity that corresponds to that corporeality of kinetic energy with the given mass. Try information technology and run into what y'all get, I'll check your piece of work if you desire.

OpenStudy (anonymous):

well i used the kinetic free energy formula and then i did the potential free energy and subtract them. when it comes to the velocity i dont know what to do

OpenStudy (whpalmer4):

\[KE = \frac{1}{2}mv^2\]Solve that equation for \(v\), then plug in the numbers.

OpenStudy (anonymous):

and so it would be v= ke/2*m?

OpenStudy (whpalmer4):

\[KE = \frac{ane}{2}mv^2\]Multiply both sides by 2\[2KE = mv^2\]Carve up both sides by \(m\)\[\frac{2KE}{grand}=v^2\]Take square root of both sides\[\sqrt{\frac{2KE}{k}}=5\]

OpenStudy (anonymous):

with the respond that i got when i subtracted KE and PE right?

OpenStudy (whpalmer4):

aye. use the remaining KE

OpenStudy (whpalmer4):

which by my figures is about 14979.ix kg thousand^two/s^two

OpenStudy (bearding):

yes that is what i got

OpenStudy (whpalmer4):

skilful!

OpenStudy (bearding):

four.97595217019 this is what i got as my final respond

OpenStudy (whpalmer4):

Yep, almost that :-)

OpenStudy (anonymous):

okay thank u

OpenStudy (anonymous):

how nigh when it has degrees?

OpenStudy (whpalmer4):

not reasonable to put that many significant digits in the answer, yet — we've just got 3 significant figures in our measurements of mass, speed, height...

OpenStudy (whpalmer4):

how nigh when what has degrees?

OpenStudy (anonymous):

yeah i got that answer correct simply i dont know how to solve this problem at present because information technology has degrees

OpenStudy (whpalmer4):

I don't see annihilation in degrees in this trouble...

OpenStudy (bearding):

hold upward

OpenStudy (anonymous):

a 77.0 kg woman slides downwardly a 42.vi m long waterslide inclined at 42.iii degress. at the bottom she is moving 20.three chiliad/southward . how much piece of work did friction on the adult female? 42.6 is not her starting height

OpenStudy (whpalmer4):

|dw:1371843934540:dw| use trig to detect the height

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A 1210 Kg Roller Coaster,

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