So, we all have to do this homework, the only thing is, I've forgotten how to do it.

Here's an example of a question:

10. A 5.0kg ball is halfway down a slope. At this point it has 40 joules of kinetic energy. At the bottom of the hill it has 150 joules of kinetic energy. By how much has its velocity changed?

So I have to do this with 2 separate equations, one for halfway and one for the bottom.

This is the formula:

KE=1/2mv[squared], so kinetic energy = 1/2 of mass x velocity squared.

So this is 40 = .5 x 5 x v x v (if 'v' is velocity) for the halfway problem.

Anyone got any ideas? I need to work out what 'v' is.

Thanks!
Last edited by BGSM at Apr 2, 2008,
Thanks, but how did you get to that? I've got more, I just need to know how you did it. I hate cheating, but I hate not having a teacher here.
Quote by BGSM
So, we all have to do this homework, the only thing is, I've forgotten how to do it.

Here's an example of a question:

10. A 5.0kg ball is halfway down a slope. At this point it has 40 joules of kinetic energy. At the bottom of the hill it has 150 joules of kinetic energy. By how much has its velocity changed?

So I have to do this with 2 separate equations, one for halfway and one for the bottom.

This is the formula:

KE=1/2mv[squared], so kinetic energy = 1/2 of mass x velocity squared.

So this is 40 = .5 x 5 x v x v (if 'v' is velocity) for the halfway problem.

Anyone got any ideas? I need to work out what 'v' is.

Thanks!

I think you need to find the change in velocity for halfway down the hill and at the bottom of the hill

Halfway down, the velocity is 2. 40 = 1/2 * 5 * v^2

20 = 5*v^2
4 = v^2
square root of four is two, so the velocity is 2

At the bottom, you get 150 = 1/2 * 5 * v^2

75 = 5v^2
15 = v^2
V = the square root of 15, which is about 3.8

So 3.8 minus 2 is a change of about 1.8. I think that's the answer.

I messed up the arithmetic. Check a few posts down.
<Han> I love Hitler
Last edited by HaKattack at Apr 2, 2008,
you figure out what v is for halfway by setting .5mv2 = 40. then you figure out what v is for the bottom by setting .5mv2 = 150. then you take the difference
Quote by BGSM
So, we all have to do this homework, the only thing is, I've forgotten how to do it.

Here's an example of a question:

10. A 5.0kg ball is halfway down a slope. At this point it has 40 joules of kinetic energy. At the bottom of the hill it has 150 joules of kinetic energy. By how much has its velocity changed?

So I have to do this with 2 separate equations, one for halfway and one for the bottom.

This is the formula:

KE=1/2mv[squared], so kinetic energy = 1/2 of mass x velocity squared.

So this is 40 = .5 x 5 x v x v (if 'v' is velocity) for the halfway problem.

Anyone got any ideas? I need to work out what 'v' is.

Thanks!

The guys above me have the right idea, also one of them messed up the arithmetic I think.

You have the formula: K = .5mv^2

So just plug stuff into it, so lets adjust our formula:

2K = mv^2

2K/m = v^2

Sqrt(2K/m) = v

Now just plug stuff in, you got all the info needed, so for the first point you get:

Sqrt(2*40/5) = v

v = 4 <---- this is your V1

Same thing for the second point,

Sqrt(2*150/5) = v

v = 7.75 <---- this is your V2

Now the difference is V2 - V1 = 7.75 - 4 = 3.75 m/s
Quote by Vasa
The guys above me have the right idea, also one of them messed up the arithmetic I think.

I did I feel dumb.

I divided by two to take away the 1/2 multiplication. Silly mistake, it's 3 in the morning.

This guy has the right answer.
<Han> I love Hitler
Last edited by HaKattack at Apr 2, 2008,
Thanks heaps, everyone. I was reading HaK's and thought that it looked wrong when dividing away the half
For anyone wondering, this is what I've done:

40 = .5 x 5 x v x v
80 = 5 x v[squared]
16 - v[squared]
v = 4.

Sweet, thanks.