# Work/Energy equation on inclined plane

• shaunchattey
In summary, Shaun struggles to work out how to calculate the resisting force on his bike when starting from a rest position. He eventually finds help from his classmates, but is frustrated because he can't work out the resistance without knowing the velocity or the energy.
shaunchattey
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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From what I could find 1 in 30 grade is 3.75 a degree decline.

Check that. I think you worked out the slope in % not degrees.

CWatters said:
Check that. I think you worked out the slope in % not degrees.

h=1 s=30
(h=1*4) /(s=30*4) = 4/120 = height of 4m

y4/x120 = 0.03 inv Tan(0.03) 1.7 degrees?

shaunchattey said:
h=1 s=30
(h=1*4) /(s=30*4) = 4/120 = height of 4m

y4/x120 = 0.03 inv Tan(0.03) 1.7 degrees?
There is no point in calculating the angle. You are given the sine of the angle (1/30), and you will only be needing trig functions of the angle. They can all be determined directly from the sine without ever knowing the angle in degrees. This will make for less work as well as improving precision.
Good technique is to work the problem entirely symbolically, ##\theta## for the angle etc., until you have a formula for the answer. Only then should you plug in numbers.

haruspex said:
There is no point in calculating the angle. You are given the sine of the angle (1/30), and you will only be needing trig functions of the angle. They can all be determined directly from the sine without ever knowing the angle in degrees. This will make for less work as well as improving precision.
Good technique is to work the problem entirely symbolically, ##\theta## for the angle etc., until you have a formula for the answer. Only then should you plug in numbers.

My issue wasnt so much that but I don't know how to work out the resistance without the velocity, and I can't find the velocity without the resistance, and I am unsure how to deal with the energy situation without one of those in this situation. I've been over this same question now so many times I've lost perspective on it.

*It would help if someone could show me their working on this question.

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shaunchattey said:
My issue wasnt so much that but I don't know how to work out the resistance without the velocity, and I can't find the velocity without the resistance, and I am unsure how to deal with the energy situation without one of those in this situation.
You don't need velocity as the resistive force is constant.
The work done by the resisting force must be equal to the initial energy, so:

mgh = Fd

where d = 120m + 40m, and F is the only unknown

shaunchattey said:
My issue wasnt so much that but I don't know how to work out the resistance without the velocity.
You didn't mention that in the OP.
shaunchattey said:
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)
why did you switch from sin to cos?

shaunchattey said:
h=1 s=30
(h=1*4) /(s=30*4) = 4/120 = height of 4m
y4/x120 = 0.03 inv Tan(0.03) 1.7 degrees?

As haruspex said, you don't actually need the angle but it would be a lot simpler to just do

Angle = Sin-1(1/30) = 1.9 degrees.

billy_joule said:
You don't need velocity as the resistive force is constant.
The work done by the resisting force must be equal to the initial energy, so:

mgh = Fd

where d = 120m + 40m, and F is the only unknown

Billy joule sorted it for me, thanks guys, I've been struggling on this one for a long time.

billy_joule
haruspex said:
You didn't mention that in the OP.

why did you switch from sin to cos?
Because I'm stupid and it was late!
Yeah I should have been more clear but I actually wasnt sure about the grades either(obviously) but you guys cleared both up.
You have no idea how much I got stuck on this question, I've never had to ask for help on a physics question before I just couldn't figure out what I was supposed to do despite having done many work energy equations before! I think it was the fact I read 'constant resistance' but I didn't take it into account, I kept trying to figure out the varied resistance like an idiot.

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## 1. What is the Work/Energy equation on an inclined plane?

The Work/Energy equation on an inclined plane is a mathematical representation of the relationship between work, energy, and the angle of inclination of the plane. It is expressed as W = Fd cosθ, where W is work, F is the applied force, d is the displacement of the object, and θ is the angle of inclination.

## 2. How does the angle of inclination affect the Work/Energy equation on an inclined plane?

The angle of inclination directly affects the amount of work done on an object on an inclined plane. As the angle of inclination increases, the amount of work done also increases, because the force required to move the object along the inclined plane increases with the angle.

## 3. Can the Work/Energy equation on an inclined plane be used to calculate the force on an object?

Yes, the Work/Energy equation on an inclined plane can be rearranged to solve for force (F = W / d cosθ). This equation can be particularly useful when the force applied to the object is unknown, but the work and displacement are known.

## 4. Is the Work/Energy equation on an inclined plane affected by friction?

Yes, the Work/Energy equation on an inclined plane takes into account the effects of friction. The work done on an object on an inclined plane is equal to the force applied times the distance moved in the direction of the force, minus the work done by friction (W = Fd cosθ - μFd cosθ).

## 5. Can the Work/Energy equation on an inclined plane be used for objects of any shape or size?

Yes, the Work/Energy equation on an inclined plane can be applied to objects of any shape or size, as long as the force and displacement are measured along the direction of motion. However, the equation assumes that the force is applied at a constant angle and that there are no external forces acting on the object.

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