Interesting and annoying power problem

1. Sep 29, 2004

diegojco

I have a trouble with this problem:

Suppose you have a particle with mass 30Kg, this particle is in influence of a force F that acts along the x axis and F=6+4x-3(x^2) and at x=0 the particle has 0m/s speed. Find the total work made by F between x=0 and x=3, then find the power given to the particle when x=3. (answers= 9 Joules and 22 Watts)

Well the first thing is easy, only I have to integrate F between 0 and 3. but the second question I can't see what's to be done. I try to use the fact that if I have the work I must get the final time but I don't have an expression for the trayectory.

Another way I have tried is to get the function of position by the fact that

F=ma, a=F/m dv/dt=(6+4x-3(x^2))/m

this is a partial differential equation involving the second derivative of a function u(x,t) and the general solution I get is:

u(x,t)=a(x)+b(x)t+(t^2)((6+4x-3(x^2))/2m)

but I don't have the sufficient conditions to get a good boundary problem or initial problem to separate a particular solution.

My questions are: what conditions could I put, or what other way, easier, could I take to solve this?

2. Sep 29, 2004

photon_mass

mdv/dt = 6 + 4x -3x^2 is a seperable second order differential equation.

problem: there are no v terms in the force only x terms.
solution: note that F is a position dependent force.

problem: can't find a term for time.
solution: is time really nescessary for finding power?

this is a much better way to formulate the problem. nevermind about parioal derivatives and general solutions and whatever else. you don't need all that for this.

3. Sep 29, 2004

krab

By conservation of energy, the work done is equal to the change in kinetic energy, which is 1/2mv^2. Use your answer to the first part and solve for v. Power is Fv.

4. Sep 29, 2004

diegojco

Thanks

Thanks for the thinking. Another problem that is easier than i have wonder.

5. Sep 29, 2004

photon_mass

power = work/time
=((force*distance)/time
= force* (distance/time)
= force*velocity

to find the velocity

dv/dt = dv/dx (dx/dt) = dv/dx *v from the chain rule.

vdv = Fdx

perform the integral. done.