interesting and annoying power problem

[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?

[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.

[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.

[diegojco]

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

[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.