Change of kinetic energy problem

[marjuna]

Hi Forum :)

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

Homework Statement

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?

Homework Equations

KE=0.5*m*V^2

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.

Thanks in advance!

 

[tiny-tim]

welcome to pf!

hi marjuna! welcome to pf!

you have d(mv²/2)/dt = 2v² …

ok, now separate the variables ​

 

[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?

 

[tiny-tim]

hi marjuna!

d(mv²/2) = -2v² dt

integrate both sides:

that's not possible!

you have to separate the variables …

see http://en.wikipedia.org/wiki/Separation_of_variables ​

 

[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?

 

[tiny-tim]

two ways:

i] treat v² as the variable, intead of v

ii] write d(v²) = 2v dv

 

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

 

[tiny-tim]

Write 1/(-2v²) * d(mv²/2) = dt as

1/(-4v²/m) d(v²) = dt

no, d(v²)/v² = d(ln(v²)) …

try substitution if you can't see it immediately ​

 

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

 

[tiny-tim]

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)?

well, yes, you do have to keep the m/2 also

write d(mv²/2) as mv dv..

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

that's right!

and now integrate …​

 

[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?

 

[tiny-tim]

looks good!

 

[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!