Joey S.
asked • 03/29/20As shown in the diagram below, a 725 kg roller-coaster is traveling at a velocity of 65.0 m/s when it is 85.0 m from the ground. By the time the coaster reached a position that was 24.0 m from the ground the coaster lost 368,050 J of energy.
Determine the velocity of the coaster, when it is 24.0 m from the ground.
Jared B. answered • 03/29/20
5+ yrs Experience: Math and Science Tutor
Find initial and final KE values at top and bottom heights.
Initial energy:
KE=1/2mvi2 + mgh
KE=m(1.2v2+gh)
KE=725kg(.5*652+9.8*85m)
KE=1833525J
Final energy:
KE=mgh=
KE=725*9.8*24
KE=170520
NRG=initial - final- loss
=1833525J-170520J-368050J
=1294955J
use thus value for KE in the equation above to solve for v
1294955J=.5(725)v2
v=60m/s
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Joey S.
Thank you. It also asks. Use 3-significant digits in your answer and don't forget your datum line. so 59.8 would be 3 sig figs but I'm not sure what it means for the Datum line?03/29/20