# Maximum vertical height

1. Dec 14, 2004

### Arty

Determining maximum vertical height

A block slides down a curved frictionless track and then up an inclined plane. The mass of the block is the point mass m. The coefficient of friction between the block and the incline is μk = 0.30 Use energy methods to find the maximum vertical height up the plane that the block reaches for θ = 45°, and h = 0.65 m.

Mgymax = mgh-Wfk
Mgymax = mgh- μkCos θ ∆x
ymax = h- μkCos θ ∆x

This is about how far I have come. Not sure if it is right though.

I am not sure if ∆x should actually be ∆y or if either should be in the problem.

I am seeking a direction on what to do next to complete this problem. I appreciate any help.

Last edited: Dec 15, 2004
2. Dec 15, 2004

### Nereid

Staff Emeritus
It seems that something is missing from your description of the problem ... presumably $$\theta$$ is the angle of the inclined plane to the vertical (or is it horizontal?); but what's h?

3. Dec 15, 2004

### marlon

Well i suppose that theta is the angle of the incline with the horizontal. Let's suppose that h is the vertical height from which we start on the curved trajectory...

At the point where the incline starts we have from energy-conservation :

$$mgh = \frac{mv^2}{2}$$

Then we move up the plane until vertical distance $$h_{max}$$

energyconservation yields :
$$\frac{mv^2}{2} = mgh_{max} - {\mu}N{\Delta p}$$ Where the delta p denotes the travelled distance of the object ON the inclined plane. ofcourse, using trigon. we know that $$h_{max} = \Delta p * sin(45)$$ so we can eliminate delta p and write all as a function of h_max...

But personally i think that there should be a least one parameter that you can vary in order to solve the extremum-problem...you sure about this question...

regards
marlon

4. Dec 15, 2004

### Tide

Not to pick nits, but ...

Is that vertical height as opposed to horizontal height?