# Conservation of energy/inclined planes

1. Aug 7, 2013

### Alyssa Jesse

I am currently taking a 1st year introductory physics paper at university. I don't have a strong background in maths or science.

1. The problem statement, all variables and given/known data
A skateboarder with a total mass of 65kg is skating on a half-pipe ramp, as shown above. When he is at the bottom of the ramp he is travelling at 12m/s. Using the principles of energy conservation, calculate how high (h) up the other side of the ramp he travels? (Assume a frictionless ride)

I'm feeling stuck as to where to start. I guess this has something to do with inclined planes, and the skateboarder would reach the same height on the opposite ramp as the height that he started at, but as the problem doesn't give me the height, I don't know how to work it out.

I'm wondering if kinetic energy would be relevant, as the formula is KE=1/2mv^2, and I have the mass and velocity?

Help would be much appreciated!

2. Aug 7, 2013

### Curious3141

You haven't attached any image of the problem.

3. Aug 7, 2013

### CAF123

When the skater is at the bottom of the ramp, all his initial potential energy has been converted to kinetic energy. Express this mathematically and you will be able to find the height relative to the base of the ramp with which he started from.

4. Aug 7, 2013

### Staff: Mentor

Sure. See the problem statement: "Using the principles of energy conservation"
At the bottom of the half-pipe, the skateboarder has kinetic energy.
At the highest point he reaches, which type of energy does he have? How do you calculate this?
How can you use energy conservation to relate those two points?

5. Aug 7, 2013

### Alyssa Jesse

Here is the image.

I worked out

KE= 1/2me^2
KE= 1/2(650*12m/s2)
KE= 46800j

Have I done this wrong? How does that equal the height of the ramp?

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Last edited: Aug 7, 2013
6. Aug 7, 2013

### Alyssa Jesse

At the height of the ramp would he have gravitational potential energy?

Last edited: Aug 7, 2013
7. Aug 7, 2013

### CAF123

Recalculate KE - the mass is given as 65kg and everything has the right units to yield a KE in SI units. That cannot equal the height of the ramp based on, besides anything else, dimensional arguments.
You are correct that at the top of the ramp the skater possesses gravitational potential energy. Now relate the energy at the top of the ramp to the energy at the bottom.

8. Aug 7, 2013

### Alyssa Jesse

The formula that I have for GPE is
GPE= weight*height

But because I don't have the height, I can't see how to use this, and how to use energy conservation to relate the two points.

9. Aug 7, 2013

### CAF123

Is height not what you are solving for?

Have you covered conservation of mechanical energy yet? Write the equation for energy when the skater is at the top of the ramp. Write the equation for energy when the skater is the bottom of the ramp. The energy at the top is converted to the energy at the bottom.

10. Aug 7, 2013

### Alyssa Jesse

So would it be -

GPE= mgh
h= GPE/mg
h= 4680/(65*10) = 7.2m?

11. Aug 7, 2013

### CAF123

That is the correct answer, but do you understand why you have GPE = (1/2)mv2?

12. Aug 7, 2013

### Alyssa Jesse

Conservation of energy? Energy cannot be created or destroyed, it may be transformed from one form into another, but the total amount of energy never changes? So in a closed system gravitation potential energy must equal kinetic energy and vice versa?

13. Aug 7, 2013

### CAF123

GPE need not equal KE at all times, but the sum of them is always constant. In your case, at the top of the ramp, all the skater's energy was GPE (KE = 0). At the bottom, all the energy was in KE (GPE = 0). At an intermediate stage between top of ramp and bottom, the energy of the skater consists of GPE and KE.

The closed system was explicit in the condition at the end of the question - 'assume frictionless ride' so there is no energy dissipation.