- #1

gelfand

- 40

- 3

[Mod note: Thread moved from General Discussion forum, so no homework template is shown]

So - I've no idea where to write this but...

I was looking at this post : https://www.physicsforums.com/threads/car-rolling-down-a-hill.302812/

And started to answer it as it was similar to some stuff that I had been doing, and was curious whether I was understanding it.

Also - the thread *kinda* got left for dead, there are no real conclusions for someone to learn from there even though it comes up in searches etc.

I'm not sure what's suggested with this kinda thing?

This is what I was *going* to reply with before I realized the thread was closed >.<

So my working was :I'm not sure if this is poor etiquette but I found this post and did some

working that I would like to check (and perhaps it's best if it's contained

within this thread for future learners)

For the first part about the speed at the bottom -

I note that I have an inclined plane (as said), it's angle and the length that

the car rolls down.

##l = ## length, and this is the hypotenuse for this triangle.

Using this I have the initial height of the car as ##h = l \times \sin(\theta)##.

From here I can use conservation of energy as

$$

W + PE_0 + KE_0 =

PE_f + KE_f + \text{Energy(Lost)}

$$

Where ##W##, ##\text{Energy(Lost)}##, ##KE_0## and ##PE_f## are all zero. ##PE_f## is zero

as at the bottom of the hill I'm considering the potential energy (due to

gravity) to be zero.

This gives me

$$

PE_0 = KE_f

$$

Which I can write as

$$

mgl \sin(\theta) = \frac{1}{2}mv^2

$$

Dividing by ##m## and rearranging gives

$$

v = \sqrt{

2 l g \sin(\theta)

}

$$

The acceleration down the hill I'm not so sure about, but I think that I have

the force of gravity pulling the car down, so this should be related to the

acceleration.

in the direction of the hill the force of gravity is

$$

F_{grav} = mg \sin(\theta)

$$

As there is no resistance I would answer with this as the acceleration.

The time that it takes for the car to reach would be dependent on my answer for

acceleration. If as I've suggested it's constant then I'm able to use the

equations of motion$$

x(t) = x_0 + v_0 t + \frac{1}{2}at^2

$$

$$

v(t) = v_0 + at

$$

$$

v^2 = v_0^2 + 2a(x - x_0)

$$

Here I would use the second one as it's purely in ##v, a## and I have this

information. Using the found expression for velocity I have

$$

\sqrt{

2 l g \sin(\theta)

} = 0 + at

$$

Dividing through by the expression found for ##a## gives

$$

\frac{\sqrt{2 l g \sin(\theta)}}{a} = t

$$

Hope this makes sense

So - I've no idea where to write this but...

I was looking at this post : https://www.physicsforums.com/threads/car-rolling-down-a-hill.302812/

And started to answer it as it was similar to some stuff that I had been doing, and was curious whether I was understanding it.

Also - the thread *kinda* got left for dead, there are no real conclusions for someone to learn from there even though it comes up in searches etc.

I'm not sure what's suggested with this kinda thing?

This is what I was *going* to reply with before I realized the thread was closed >.<

So my working was :I'm not sure if this is poor etiquette but I found this post and did some

working that I would like to check (and perhaps it's best if it's contained

within this thread for future learners)

For the first part about the speed at the bottom -

I note that I have an inclined plane (as said), it's angle and the length that

the car rolls down.

##l = ## length, and this is the hypotenuse for this triangle.

Using this I have the initial height of the car as ##h = l \times \sin(\theta)##.

From here I can use conservation of energy as

$$

W + PE_0 + KE_0 =

PE_f + KE_f + \text{Energy(Lost)}

$$

Where ##W##, ##\text{Energy(Lost)}##, ##KE_0## and ##PE_f## are all zero. ##PE_f## is zero

as at the bottom of the hill I'm considering the potential energy (due to

gravity) to be zero.

This gives me

$$

PE_0 = KE_f

$$

Which I can write as

$$

mgl \sin(\theta) = \frac{1}{2}mv^2

$$

Dividing by ##m## and rearranging gives

$$

v = \sqrt{

2 l g \sin(\theta)

}

$$

The acceleration down the hill I'm not so sure about, but I think that I have

the force of gravity pulling the car down, so this should be related to the

acceleration.

in the direction of the hill the force of gravity is

$$

F_{grav} = mg \sin(\theta)

$$

As there is no resistance I would answer with this as the acceleration.

The time that it takes for the car to reach would be dependent on my answer for

acceleration. If as I've suggested it's constant then I'm able to use the

equations of motion$$

x(t) = x_0 + v_0 t + \frac{1}{2}at^2

$$

$$

v(t) = v_0 + at

$$

$$

v^2 = v_0^2 + 2a(x - x_0)

$$

Here I would use the second one as it's purely in ##v, a## and I have this

information. Using the found expression for velocity I have

$$

\sqrt{

2 l g \sin(\theta)

} = 0 + at

$$

Dividing through by the expression found for ##a## gives

$$

\frac{\sqrt{2 l g \sin(\theta)}}{a} = t

$$

Hope this makes sense

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