# Conservation of mechanical energy (low difficulty)

1. Feb 11, 2009

### Oscar Wilde

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

There is a rollercoaster. A car starts 35m "high" and descends to the ground. A.) What is the velocity at this point? The car continues an goes 28m above the ground. B.) calculate its velocity at this point. C.) If the car descends another 13m from this point, what is the velocity

Ignore Friction
2. Relevant equations

mgy=1/2mv^2

3. The attempt at a solution

using mgy= 1/2mv^2 i got the answer to a.), which is 26 m/s. But after this I became flustered and I do not know what to do to get its velocity if it goes 28m higher. I understand c. but I need the velocity at point B to get it!

2. Feb 11, 2009

### Oscar Wilde

I'm sorry I just figured it out for B.

I set 1/2mv^2=1/2mv^2+mgy and found my second velocity, which comes out to 11.7 roughly when using g=9.8

I would appreciate if someone would confirm my findings

I then set up the (square root of(1/2mv^2 + mgy-mgy)/.5)=v and came out with 20 m/s for C.

3. Feb 11, 2009

You're on the right track, think about conservation of energy
$$\Delta U = m g \Delta y$$
$$\Delta K = \frac{1}{2} m(v_f ^2-v_i^2)$$
Thus $$\Delta K + \Delta U = 0$$
Thus $$-m g \Delta y +\frac{1}{2} m v_i^2= \frac{1}{2} m v_f^2$$

4. Feb 12, 2009

### jumbogala

I also got 11.7 for part B.