How to balance gravitational energy and kinetic energy?

[Eclair_de_XII]

Homework Statement

"In Fig. 8-33, a runaway truck with failed brakes is moving downgrade at $130\frac{km}{h}$ just before the driver steers the truck up a friction-less emergency escape ramp with an inclination of $θ=15°$. The truck's mass is $1.2⋅10^4kg$. What minimum length L must the ramp have if the truck is to stop momentarily along it?

Homework Equations

$v=130\frac{km}{h}=\frac{325}{9}\frac{m}{s}$
$m=1.2⋅10^4kg$
$θ=15°$
$K=\frac{1}{2}mv^2$
$-K=-Fy=-(mgsinθ)(y)$
$y=Lsinθ$

The Attempt at a Solution

First, I figured that the gravitational work must equal to the work produced by the truck and its velocity. And gravity does work only in the vertical direction, which is why I'm expressing gravitational work in terms of the force and the vertical distance $y$. This is what I did, and it isn't giving me the correct answer.

$K=\frac{1}{2}mv^2$
$-K=-Fy=-(mgsinθ)(y)$
$K=Fy=(mgsinθ)(y)=(mgsinθ)(Lsinθ)$
$(mgsinθ)(Lsinθ)=(\frac{1}{2}mv^2)$
$(L)(g)(sin^2θ)=(\frac{1}{2}v^2)$
$L=(\frac{1}{2g})(v^2)(csc^2θ)$

Evaluating this gives me: $L=993.2m$, which is not the answer from the back of the book. I feel like there's something I'm not understanding. Can someone help me? Thanks.

 

[ehild]

Homework Statement

"In Fig. 8-33, a runaway truck with failed brakes is moving downgrade at $130\frac{km}{h}$ just before the driver steers the truck up a friction-less emergency escape ramp with an inclination of $θ=15°$. The truck's mass is $1.2⋅10^4kg$. What minimum length L must the ramp have if the truck is to stop momentarily along it?

Homework Equations

$v=130\frac{km}{h}=\frac{325}{9}\frac{m}{s}$
$m=1.2⋅10^4kg$
$θ=15°$
$K=\frac{1}{2}mv^2$
$-K=-Fy=-(mgsinθ)(y)$
$y=Lsinθ$

The Attempt at a Solution

First, I figured that the gravitational work must equal to the work produced by the truck and its velocity. And gravity does work only in the vertical direction, which is why I'm expressing gravitational work in terms of the force and the vertical distance $y$. This is what I did, and it isn't giving me the correct answer.

What is the potential energy at high y?

 

[Eclair_de_XII]

It's not $(mg)(sinθ)(y)$, is it?

 

[ehild]

It's not $(mg)(sinθ)(y)$, is it?

No. How much work you have to do if you raise a mass m to height y? Does the work depend on the path along the mass moves?

 

[Eclair_de_XII]

How much work you have to do if you raise a mass m to height y?

The formula for work is just force dot distance, so $W=(F)(y)(cosθ)$, depending on which angle we're using.

Does the work depend on the path along the mass moves?

It does not, I believe.

 

[ehild]

The formula for work is just force dot distance, so $W=(F)(y)(cosθ)$, depending on which angle we're using.
It does not, I believe.

You said y was the vertical component of the displacement, y=Lsin(θ). What is the direction of the force?

 

[Eclair_de_XII]

What is the direction of the force?

Isn't it forty degrees? Additionally, I feel like I may have to express velocity in terms of its x- and y-components. Am I wrong?

 

[ehild]

Isn't it forty degrees? Additionally, I feel like I may have to express velocity in terms of its x- and y-components. Am I wrong?

What force acts on the truck along the upward ramp?

 

[Eclair_de_XII]

Gravity.

 

[ehild]

Gravity.

Yes. What is the direction of gravity?

 

[Eclair_de_XII]

It's always pointing down... unless you shift the axes of the FBD, I think.

 

[ehild]

It's always pointing down... unless you shift the axes of the FBD, I think.

Yes, it is vertical, pointing down. What is the potential energy of a body of mass m at height y?

 

[Eclair_de_XII]

Is it $U=(F_G)(y)$? I've also said that this is equal to the quantity $[(mg)(sin\theta)]⋅[(L)(sin\theta)]$. This isn't correct, is it?

 

[ehild]

Is it $U=(F_G)(y)$? I've also said that this is equal to the quantity $[(mg)(sin\theta)]⋅[(L)(sin\theta)]$. This isn't correct, is it?

Yes, the potential energy is mgy and y = Lsin(θ). You were not correct.

 

[Eclair_de_XII]

Yes, the potential energy is mgy and y = Lsin(θ). You were not correct.

So the force of gravity is not $(mg)(sin\theta)$ when doing energy calculations? In any case, this gave me the right answer.

$L=(\frac{1}{2g})(v^2)(csc\theta)=(\frac{1s^2}{19.6m})(\frac{105625}{81}\frac{m^2}{s^2})(csc15°)=257m$

Thanks. I appreciate it.

 

[ehild]

So the force of gravity is not $(mg)(sin\theta)$ when doing energy calculations? In any case, this gave me the right answer.

$L=(\frac{1}{2g})(v^2)(csc\theta)=(\frac{1s^2}{19.6m})(\frac{105625}{81}\frac{m^2}{s^2})(csc15°)=257m$

Thanks. I appreciate it.

The force of gravity is mg. Its component along the slope is mgsin(θ), downward. You can also calculate the change of potential energy along the slope. It is also mgsin(θ) L.

 

[haruspex]

So the force of gravity is not (mg)(sinθ) when doing energy calculations?

The angle of interest is the angle between the direction of the force and the direction of the distance you are multiplying by. If the angle between the two is θ then the work is Fd cos(θ).
Gravity is vertical. If you want to know the height the truck goes, that is a vertical distance. The angle between the two is zero.
If you want to know the distance up a ramp at angle θ, the angle between that and g is π/2-θ.

 

[Eclair_de_XII]

You can also calculate the change of potential energy along the slope. It is also mgsin(θ) L.

I see.

The angle of interest is the angle between the direction of the force and the direction of the distance you are multiplying by. If the angle between the two is θ then the work is Fd cos(θ).

I will have to remember that in the future. Thanks.