Mechanics work-energy question

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SUMMARY

The discussion revolves around a mechanics work-energy problem involving a cyclist descending a slope of 1 in 30. The cyclist, with a total mass of 72 kg, freewheels 120 m down the slope before traveling an additional 40 m on a horizontal road. The participant calculated the gravitational potential energy (MGH) to be 4923 J but struggled with determining the velocity and resisting force. Clarification on the slope's grade and the use of work-energy equations were emphasized as critical for solving the problem.

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  • Understanding of gravitational potential energy (MGH)
  • Knowledge of slope calculations and trigonometry
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  • Ability to apply Newton's laws of motion
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  • Learn how to calculate resisting forces in motion problems
  • Explore the relationship between slope angles and trigonometric functions
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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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