Homework Help: How to balance gravitational energy and kinetic energy?

1. Oct 2, 2016

Eclair_de_XII

1. The problem statement, all variables and given/known data
"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?

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

3. 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.

2. Oct 2, 2016

ehild

What is the potential energy at high y?

3. Oct 2, 2016

Eclair_de_XII

It's not $(mg)(sinθ)(y)$, is it?

4. Oct 2, 2016

ehild

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?

5. Oct 2, 2016

Eclair_de_XII

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.

6. Oct 2, 2016

ehild

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

7. Oct 2, 2016

Eclair_de_XII

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?

8. Oct 2, 2016

ehild

What force acts on the truck along the upward ramp?

9. Oct 2, 2016

Eclair_de_XII

Gravity.

10. Oct 2, 2016

ehild

Yes. What is the direction of gravity?

11. Oct 2, 2016

Eclair_de_XII

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

12. Oct 2, 2016

ehild

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

13. Oct 2, 2016

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?

14. Oct 2, 2016

ehild

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

15. Oct 2, 2016

Eclair_de_XII

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.

16. Oct 2, 2016

ehild

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.

17. Oct 2, 2016

haruspex

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.

18. Oct 3, 2016

Eclair_de_XII

I see.

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