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## Homework Statement

A roller coaster reaches the top of the steepest hill with a speed of 7.0 km/h. It then descends the hill, which is at an average angle of 43° below horizontal and is 46.0 m long. What will its speed be when it reaches the bottom? Assume the coefficient of kinetic Friction is 0.12.

## Homework Equations

Ff= U•Fn

F=ma

Vf ^2 - Vo^2 = 2ad

## The Attempt at a Solution

Ff=U•Fn

Ff= U•mgcos(43)

F=ma

Mgsin(43) - Umgcos(43) =ma

The mass cancels out here so ,

gsin(43) - Ugcos(43) = a

5.82 = a

Vf= (square roof of>>) 2ad + vo ^2

Vf = (square roof of >> ) 2 • 5.82 • 0.046 + 7.0^2

Vf = 7.04 km/h

The answer should be 83.6 km/h

Where did I go wrong