A roller coaster reaches the top of the steepest hill with a speedof 6.0 km/h. I

Question

A roller coaster reaches the top of the steepest hill with a speedof 6.0 km/h. It then descends the hill, which is at an average angle of 45 degrees and it 45.0m long. Estimate its speed when it reaches the bottom. Assume coefficient of kinetic friction is .18

this is what ive done so far... Fgx=mgsin45
Fgy=mgcos45=Fn Vo=6.0km/h x 1000m/1km x 1h/3600s = 1.67m/s
EF=Fgx-Ff=ma
mgsin45-umgcos45=ma
g(sin45-ucos45)=a
5.68 m/s^2=a

this is where im not sure...

V^2=Vo^2+2a(x-xo) V^2= (1.67)^2+2(5.68)(45) V^2= 513.99 V=22.67 m/s Does that look right? this is what ive done so far... Fgy=mgcos45=Fn Vo=6.0km/h x 1000m/1km x 1h/3600s = 1.67m/s V^2=Vo^2+2a(x-xo) V^2= (1.67)^2+2(5.68)(45) V^2= 513.99 V=22.67 m/s Does that look right?

Explanation / Answer

Yes, you`re absolutely right.

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