# Conservation of Energy Help

1. Mar 17, 2014

### lostinphysics1

1. The problem statement, all variables and given/known data
I have a homework assignment that I am stuck on using the attached table.

Problem 1: I have to find the velocity when given the mass and height.
Problem 2: Then I have to find the height when given the velocity and mass.

2. Relevant equations
P.E. = m*g*h
K.E. = 1/2*m*v^2

Mass = 1kg
Gravity = 9.8m/s

3. The attempt at a solution
P.E.(1) = 1kg * 9.8m/s * .5m
P.E.(1) = 4.9J

P.E.(2) = 1kg * 9.8m/s * .2m
P.E.(2) = 1.96J

K.E.(1) = 0m/s

K.E.(2) = 1/2*1kg*(.8m)^s
K.E.(2) = .32J

K.E.(1) + P.E.(1) = K.E.(2) + P.E.(2)
0m/s+4.9J = 1/2*1kg*v^2 + 0J
4.9J = .5*v^2
4.9J*.5=v^2
√2.45J=v^2
2.45 m/s = v

I am not sure about how to get the height.

Thank you for any help

#### Attached Files:

• ###### table.jpg
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2. Mar 17, 2014

### iRaid

$$KE_i+PE_i=KE_f+PE_f$$

Just use that and change out the KE's and PE's with their respective formulas.

3. Mar 17, 2014

### lostinphysics1

But that will only provide data in Joules and not a velocity or height?

Thank you

4. Mar 17, 2014

### iRaid

Think about it, the first law of energy conservation is that the total energy of an isolated system cannot change (or cannot be created nor destroyed), so whenever something happens, something else has to counteract it. If you have an initial KE and PE, you have to have a final KE and PE.

So, the equation becomes:
$$\frac{1}{2}mv^2_i+mgh_i=\frac{1}{2}mv^2_f+mgh_f$$

All the mass values cancel and you're left with:
$$\frac{1}{2}v^2_i+gh_i=\frac{1}{2}v^2_f+gh_f$$

Now all you do is plug in those values you have in the table.

5. Mar 17, 2014

### lostinphysics1

I will have to dust off my algebra book to move the variables around as the information in the chart is provided.
I will try it in the morning see what I get.

Thanks for the help!

6. Mar 18, 2014

### lostinphysics1

I emailed the teacher and asked for help but he provided answers:
Problem #1: 2.42 m/s^2
Problem #2: .259 m

When I input the values from the chart I do not get the answers he provided. So, I am not sure what I am doing wrong.

Does anyone else get the same solutions mentioned above?

7. Mar 18, 2014

### iRaid

Nope I get the same answers as him, did you even do what I said before?

8. Mar 18, 2014

### lostinphysics1

Yes, I tried but I had no way to know if the answers I got are correct so I sent the answers I had. Which as it turns out had been wrong.

But I think I finally got Problem 1.

P.E.i = 4.9J
K.E.i = 0J
P.E.f = 1.96J
K.E.f = ?

K.E.i + P.E.i = K.E.f + P.E.f
0J + 4.9J = 1/2 * 1kg * (v)^2 + 1.96J
4.9J = .5 *(v)^2
-1.96 = .5 * (v)^2
2.94 = .5 * (v)^2
/.5 = (v)^2
5.88 = (v)^2
√5.88 = √(v)^2
2.42 = V

Now to try Problem 2

Thank you

9. Mar 18, 2014

### iRaid

First one is correct, now do the same for the 2nd, but you're trying to find the final height now.

10. Mar 18, 2014

### lostinphysics1

Did Problem 2 similar to the first.

P.E.i = m*g*h
P.E.i = 1kg * 9.6m/s * .3m
P.E.i = 2.94J

K.E.i = 1/2 * m * (v)^2
K.E.i = 1/2 * 1kg * (.8m/s)^2
K.E.i = .32J

K.E.f = 1/2 * m * (v)^2
K.E.f = 1/2 * 1kg * (1.2m/s)^2
K.E.f = .72

Need P.E.f

P.E.i + K.E.i = P.E.f + K.E.f
2.94J + .32J = 1kg * 9.8m/s * h + .72J
3.26J = 1kg * 9.8m/s * h + .72J
-.72J - .72J
2.54J = 1kg * 9.8m/s * h
2.54J = 9.8 kg*m/s * h
2.54J/9.8 kg*m/s = h
.259 m = h

Seems my answer matches up with the teachers.

Thank you for the help

Hopefully this example helps others