# Homework Help: Change of kinetic energy problem

1. Feb 28, 2012

### marjuna

Hi Forum :)

This is not a specific homework problem, just something I tried to solve myself.

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

A 10 kg object is moving at 10 m/s.

The object is losing 2v^2 J of kinetic energy per second.

Determine the time it takes for the speed to decrease from v_1 = 10 m/s to v_2 = 5 m/s.

EDIT: or even better, how do I obtain an expression for velocity as a function of time?

2. Relevant equations

$$KE=0.5*m*V^2$$

3. The attempt at a solution

Alright, so the difference in kinetic energy between v_1 and v_2 is -375 J. I also noticed that the derivative of KE with respect to v is -2v^2. I somehow need to get the time variable in there.. How do I continue to solve the problem? I'm not interested in a numerical answer, more in the way of doing it.

Last edited: Feb 28, 2012
2. Feb 28, 2012

### tiny-tim

welcome to pf!

hi marjuna! welcome to pf!

you have d(mv2/2)/dt = 2v2

ok, now separate the variables

3. Feb 28, 2012

### marjuna

Thanks tiny-tim!

So you get:

d(mv2/2)/dt = -2v2

d(mv2/2) = -2v2 dt

integrate both sides: d(mv2/2) was -375 J

-375 = -2v2*t

Uhm, how exactly do I continue from here?

4. Feb 28, 2012

### tiny-tim

hi marjuna!
that's not possible!!

you have to separate the variables

5. Feb 28, 2012

### marjuna

oooh, sorry

write it as:

1/(-2v2) * d(mv2/2) = dt

THEN integrate both sides right?

But how do I integrate with respect to mv2/2 when the variable on the left side is just v?

6. Feb 28, 2012

### tiny-tim

two ways:

i] treat v2 as the variable, intead of v

ii] write d(v2) = 2v dv

7. Feb 28, 2012

### marjuna

Okay, then:

case i]

Write 1/(-2v2) * d(mv2/2) = dt as

1/(-4v2/m) d(v2) = dt

Then integrate the left side from v_1 to v_2 (substitute values) and the right side becomes t, obviously and then you've got an answer, right?

and if you go for ii] then the equation becomes:

m/-2v dv = dt

and then also integrate both sides

but one of them must be wrong since i get different expressions after integrating.. hmm

8. Feb 28, 2012

### tiny-tim

no, d(v2)/v2 = d(ln(v2)) …

try substitution if you can't see it immediately

9. Feb 28, 2012

### marjuna

Uhm, I don't get it :(

How do you get rid of the m/2 inside d(mv2/2) so you're left with just d(v2)?

But if you do it the other way:

write d(mv2/2) as mv dv..

then the equation becomes m/-2v dv = dt

is that correct? oh man I really should start polishing up those calculus skills :)

Last edited: Feb 28, 2012
10. Feb 28, 2012

### tiny-tim

well, yes, you do have to keep the m/2 also
that's right!

and now integrate …​

11. Feb 28, 2012

### marjuna

Alright then!

After integration the equation becomes:

t = -m/2 * ( ln[v_2] - ln[v_1] )

Substituting, m = 10 kg, v_1 = 10 m/s and v_2 = 5 m/s gives t ≈ 3.47 s

Seems plausible, right?

12. Feb 28, 2012

### tiny-tim

looks good!

13. Feb 28, 2012

### marjuna

Alright then.

tiny-tim, thank you a lot for your time and help, I really appreciate it I hope you have a great day!