# Help w/ conservation of mech. energy

1. Jan 23, 2005

### Femme06Fatale

Alright - i have a huge physics test on tuesday and certainly need TONS of help w/ this stuff - i have a really hard time visualizing what it is that we are calculating. So now, we were just introduced to the conservation of mechanical energy - so i was wondering if there's anyone out there that would care to help me w/ it .. like trying to understand any of it -- PLEASE :) It'd be greatly appreciated! Thanks much :)

2. Jan 23, 2005

### Sirus

You're going to have to be more specific. Mechanical energy is the sum of the total kinetic and potential energy of a system or object.

$$E_{\mbox{mechanical}}=E_{\mbox{kinetic}}+E_{\mbox{potential}}$$

Mechanical energy isn't really anything you can "understand" or visualize; just think of it as a useful term for defining the above sum.

3. Jan 23, 2005

### Femme06Fatale

The largest watermelon ever grown had a mass of 118 kh. SUppose this watermelon is exhibited on a platform 5.00 m above the ground. After the exhibition, the watermelon is allowed to slide to the ground along a smooth ramp. How high above the ground is the watermelon at the moment its kinetic energy is 4.61 kJ? -- Where do I even start?

4. Jan 23, 2005

### Sirus

We assume an isolated system (no non-conservative forces like friction), so we can say that mechanical energy is conserved. Therefore, at any point during its fall the watermelon has only as much energy as it had before it was allowed to slide to the ground. You are given kinetic energy and you have information about the mechanical energy.

Use $$E_{\mbox{potential gravitational}}=mgh$$.

5. Jan 23, 2005

### christinono

Start by finding the initial mechanical energy. At the top of the platform, if the watermelon in motionless, its total mechanical energy is the gravitational potential energy, since there is no kinetic energy. To solve the problem, remember that the total mechanical energy (sum of Ep and
Ek) always remains constant.

6. Jan 24, 2005

### Femme06Fatale

The deepest mine ever drilled has a depth of 12.3km. Suppose you drop a rock w/ a mass of 120.g down the shaft of this mine. What would the rock's kinetic energy be after falling 3.2km? What would the potential energy associated w/ the rock be at that same moment? Assume no air resistance & a constant free-fall acceleration.

So I know that the equation for ke=1/2mv^2 and I know that pe=mgh
So do I just plug in what i know for pe ... making it 120(9.81)12.3 for pe and then for ke, i have to find velocity first --- can anybody tell me if I'm thinking correctly?

7. Jan 24, 2005

### Staff: Mentor

(1) Energy is conserved. That means: $\Delta {KE} = - \Delta {PE}$.
(2) What's the change in height? (It's not 12.3 km.)
(3) The mass is 120 g. Convert to kg.