Question #148037
A rollercoaster applies 36000N of force to a 2600kg train of coaster cars for 2 seconds to launch the coaster up the first hill. The launch platform is 3 meters above the ground. How much work did the launcher do and what is the maximum height of the first hill on the rollercoaster if the coaster still has a velocity of 0.25 m/s at the top of the first hill?
1
Expert's answer
2020-12-02T09:45:26-0500
Ft=mv36000(2)=2600vv=27.69msFt=mv\to 36000(2)=2600v\\v=27.69\frac{m}{s}

W=0.5mv2=0.5(2600)(27.69)2=996757 JW=0.5mv^2=0.5(2600)(27.69)^2=996757\ J

v2u2=2gh27.6920.252=2(9.8)hh=39 mv^2-u^2=2gh\\27.69^2-0.25^2=2(9.8)h\\h=39\ m

H=39+3=42 mH=39+3=42\ m


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