How Do You Calculate Velocity and Height Using Conservation of Energy?

AI Thread Summary
The discussion focuses on solving two physics problems using the conservation of energy principle. For Problem 1, the user calculates the velocity from given mass and height, resulting in a velocity of 2.42 m/s after applying the kinetic and potential energy formulas. In Problem 2, the user successfully determines the height using the same conservation of energy approach, arriving at a height of 0.259 m, which matches the teacher's provided answers. The conversation emphasizes the importance of understanding energy conservation and correctly applying the relevant equations. Overall, the user gains clarity on both problems through community support and guidance.
lostinphysics1
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Homework Statement


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.

Homework Equations


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

Mass = 1kg
Gravity = 9.8m/s

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
 

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KE_i+PE_i=KE_f+PE_f

Just use that and change out the KE's and PE's with their respective formulas.
 
iRaid said:
KE_i+PE_i=KE_f+PE_f

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

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

Thank you
 
lostinphysics1 said:
But that will only provide data in Joules and not a velocity or height?

Thank you

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.
 
iRaid said:
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.

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!
 
lostinphysics1 said:
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!

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

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
 
lostinphysics1 said:
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

First one is correct, now do the same for the 2nd, but you're trying to find the final height now.
 
  • #10
iRaid said:
First one is correct, now do the same for the 2nd, but you're trying to find the final height now.

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
 
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