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shaunchattey

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**I've been trying to work out what I'm doing wrong on this question but I just don't get it.**

'A cyclist starting from rest freewheels for 120m down a slope of 1 in 30. At the bottom of the slope the road becomes horizontal, and the cyclist stops without using his breaks after going a further 40m. If the total mass of the bicycle and rider is 72kg, find the resisting force, assumed constant'

From what I could find 1 in 30 grade is 3.75 a degree decline.

Potential energy= (mgh)72*10*sin(3.75)(120)

Force of gravity over distance(mgs) = 72*10*cos(3.75)(120) + (72*10*40)

Force over distance vs resistance over distance = 72*10*sin(3.75)-R*(72*10*cos(3.75)*120) + (72*10*40) = Crazy wrong answer.

I've been unable to figure this one out for quite a while, I occasionally go back to it. This is just one of many incorrect answers I've ended up with.

(The answer given in the book is 18N)

I've also tried working backwards to find the forces of gravity from knowing the resistance is 18N, didn't help, ended up with ridiculously small numbers.

Help.

Shaun.

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