Physics help.?

Question

I would just like to know if I have a function of position x=3t-3t^2+t^3, with a mass m=3kg. Then all I would have to do is take my dv/dt function times my mass and then integrate it from x(0) to my X(4) to get my work done over that interval. I get 720J as my answer. Does that make since?

Answer 1

You're on the right track.

dW = Fdx = m dv/dt * (dt*v) = mv dv

W = 1/2 mv^2 evaluated at v(4) and v(0)

This result is obvious if you've heard of kinetic energy.

Answer 2

Yes, it does make sense.

There is however a better way.
Instead of integrating over force over dx,
it's possible to integrate power over time dt:

dWork = Force * dx
dWork = ma * dx
dWork = m(dv/dt) * dx
dWork = m dv dx/dt
dWork = m dv v
dWork = 1/2 m dv^2
integrating:
Work = 1/2 mV^2 - 1/2 mVo^2
This is, of course, conservation of enegry.

v = df(t)/dt = 3 - 6t + 3t^2
Work =
= 1/2 * 3 * [(3 - 24 + 48)^2 - (3)^2] =
= 1/2 *3 * [27^2 - 3^2]
= 1080 J

Answer 3

I went about it a bit differently.

Power is force times speed, and work is the integral of force times speed over the time interval.

First of all, F=m*a To find a, take the second derivative
d(v(t))/dt=a(t)
dx(t)/dt=v(t)=3-6t+3t^2

a(t)=-6+6t
since F=m*a and P=F*v
then work is the integral of the mass times
-18+54t-54t^2+18t^3
We can see by inspection that x=0 when t=0
Also, at what t is x=4?
4=3t-3t^2+t^3
or
0=-4+3t-3t^2+t^3
t=2.442 s

so the integral is from t=0 to t=2.442 times the mass
or 45 J

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