Does Velocity Remain Constant at Each Hilltop on a Frictionless Track?

[Texhno]

If a block of wood is placed on a hilly frictionless track, would the velocity stay constant at the top of each hill if the height of each hill was the same? Since as the block of wood slides down a hill, it gains kenetic energy looses potential energy, but as it rises up the hill, the PE increases and KE decreases so the velocity returns to the orginal magnitude as it was on the top of the last hill. Do I have this concept correct?

 

[TALewis]

Yes, that sounds right, so long as the track is frictionless and there are no other losses in energy anywhere else (air resistance, etc.).

 

[Texhno]

so if the block of wood went over a hill that was half the height of an original hill, then the velocity would double because not all of the KE is turned into PE. Since there is access KE, the velocity would increase by half.

am I correct?

 

[remcook]

so..what if there is a large hill where the velocity at the top is exactly zero? On the top of a a hill half the size, according to your answer, you would have velocity: 0.5*0 =0!

Try consistently applying conservation of energy: at every point the total energy is the same.

 

[Texhno]

Even though the block would gain velocity down a hill and have more velocity to attack the second hill that has half the height, the velocity is actually half the original because the hill is half the original height?

 

[Texhno]

saying that mg(.5h)=.5m(.5*v^2)
since the height is cut in half, then the KE is also cut in half?

 

[TALewis]

I'm not exactly following what you're asking, but it's just a matter of consistently applying the conservation of energy. For any position along the track, the total mechanical energy (potential + kinetic) of the system remains constant. That is to say, the following is true:

$$mgh + \frac{1}{2}mv^2 = \textrm{constant}$$

Or, between any two points 1 and 2 on the track:

$$mgh_1 + \frac{1}{2}mv_1^2 = mgh_2 + \frac{1}{2}mv_2^2$$

Try using these equations to carefully work through what you're trying to find out.

 

[Texhno]

Then I'm correct...I said "so if the block of wood went over a hill that was half the height of an original hill, then the velocity would double because not all of the KE is turned into PE. Since there is access KE, the velocity would increase by half." I think I meant to say "the velocity would double."

mgh+.5mv^2=mg.5h`+.5m2v`^2
in the second part, since height is halfed, then the velocity must double

 

[TALewis]

That's not true. You have to look at the entire situation by setting up the equation and actually solving for the variable you're interested in.

Let v1 be the velocity at the top of the hill of height h. Let v2 be the velocity on the hill of 0.5h. We want to find v2 as it relates to v1. You say v2 is double v1. Let's look:

$$\begin{align*}<br /> mgh + \frac{1}{2}mv_1^2 &= mg(0.5h) + \frac{1}{2}mv_2^2\\<br /> (0.5)mgh + \frac{1}{2}mv_1^2 &= \frac{1}{2}mv_2^2<br /> \quad \mbox{(cancel \frac{1}{2}m on both sides)}\\<br /> gh + v_1^2 &= v_2^2\\<br /> v_2 &=\sqrt{gh + v_1^2}\quad \mbox{(ans.)}<br /> \end{align}$$

As you can see, v2 depends not only on v1, but also on gh. It cannot be simply said that v2 is always twice v1; in fact, you can find that's only the case when h = 3v1^2/g .

 

[Texhno]

if this ball with a velocity of sqrt(gh+v^2) on the top of the hill rolls down that hill and rolls to a flat surface. How much force is required to stop the ball in "d" meters? I know that F=ma but there is no exceleration so what is the force required?

 

[TALewis]

You don't even need to consider the intermediate hill. Keeping the same conventions for h and v1 as before, at any time the ball has total mechanical energy given by:

$$\mbox{(mechanical energy)} = mgh + \frac{1}{2}mv_1^2$$

After rolling down to the flat surface, the ball still has this total mechanical energy. Also, if a force F acts on the ball over a distance d, then the work done by the force on the ball is F*d. This work done to stop the ball must equal the mechanical energy that the ball started with. That is:

$$Fd = mgh + \frac{1}{2}mv_1^2$$

From that, you can easily solve for F.

 

[Texhno]

Can you check if I did these problems right?

Questions are on this site: http://home.earthlink.net/~suburban-xrisis/pics01.jpg

84) PE=mgh
85) PE=KE
mgh=.5mv^2
gh=.5v^2
v=sqrt(2gh)
86) PE=mgh
PE=.5kx^2
mgh=.5k(-y)^2
87) 0.5kx^2=.5mv^2
k=(mv^2)/(x^2)
k=(mv^2)/(y^2)

 

[TALewis]

84 and 85 look right.

86: I think the problem wants you to write conservation of energy between B and C.

At B, the block is at y = 0, so there is no potential energy. It has a kinetic energy of 0.5mv^2, however.

At C, there is the energy of the spring, which you have correct as 0.5ky^2. However, the block has moved below the datum line a distance -y, so there is negative potential energy of -mgy. Putting these together, I get:

$$\frac{1}{2}mv^2 = \frac{1}{2}ky^2 - mgy$$

As for 87, you should be able to solve the above for k in terms of g, v, m, and y.