MHB Mechanics work-energy question

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A cyclist freewheels down a 1 in 30 slope for 120m before stopping on a horizontal road after 40m. The total mass of the bicycle and rider is 72kg, and the cyclist is trying to determine the resisting force, assuming it remains constant. There was confusion regarding the grade of the slope, which was calculated as approximately 3.33% or a 4-degree angle. The cyclist initially struggled with calculating velocity and resistance without knowing one of the values. Clarification on the slope's description is necessary for accurate calculations.
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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 brakes after going a further 40m. If the total mass of the bicycle and rider is 72kg, find the resisting force, assuming it to be constant throughout.

I've been teaching myself physics and maths over the last 6 months and have got stuck on a problem today.
I had trouble figuring out exactly what the grade represents (I haven't used it before) I calculated it to be 3.33% slope or possibly a 4 degree slope if it drops 1 degree every 30m.

Either way I worked out the MGH to be 72*9.81*(sin3.33 * 120) 4923 J

I know I screwed up the grade but beyond that I couldn't figure out how to calculate either the velocity or resistance without knowing one or the other.

Thanks for any help this question has been annoying me and I'm at the point where I've tried too many different things I've only confused myself.

edit: Nevermind I'll work it out myself, I've worked everything out so far. I'll go over the work energy and work against resistance equations again.
 
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You need to specify what do you mean by "1 in 30" in the first line of your question.
 
phymat said:
You need to specify what do you mean by "1 in 30" in the first line of your question.

I take that to mean that for every 30 m moved forward horizontally, the road drops 1 m vertically. Therefore the angle of inclination $\theta$ would be:

$$\theta=\arctan\left(\frac{1}{30}\right)$$
 
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