# Finding magnitude of net force

1. Jan 30, 2010

### danielatha4

1. The problem statement, all variables and given/known data
A 2.60 kilogram mass is moving in a plane, with its x and y coordinates given by x = 5.40t2 - 1.30 and y = 2.75t3 + 1.80, where x and y are in meters and t is in seconds. Calculate the magnitude of the net force acting on this mass at t = 2.10 seconds.

2. Relevant equations
Our teacher has been making us use the position update formula and the momentum principle:
$$\Delta$$r=Vavg$$\Delta$$t
$$\Delta$$P=Fnet$$\Delta$$t

3. The attempt at a solution

I can figure out the velocity at a given point in both the y and x directions by finding the derivative:
dx/dt=8.4t
dy/dt=8.25t2

But I don't know how to apply this. I really don't know where to start.

Edit: sorry, dx/dt should be 10.8t

Last edited: Jan 30, 2010
2. Jan 30, 2010

### pgardn

Well the second derivative of the position functions or the first derivative of the velocity functions should give you a(t). Then the second law F=ma. So you have then you should have the forces in the x and y direction at the time given. Add those two vectors together... pythag. theorem.

3. Jan 30, 2010

### danielatha4

Thanks for the reply. I figured out that way of solving this problem after I posted it. However, our teacher won't let us use f=ma yet, as we haven't gotten to it in our class yet.

I no longer NEED help on this problem since I have the correct answer, but does anyone know how to do it with only the position update formula and momentum principle?

4. Jan 31, 2010

### gabbagabbahey

Once you know the components of velocity, just use them to calculate the momentum at two different times, say $t=t_0$ and $t=t_0+\Delta t$, and substitute it into the momentum principle formula you gave. The smaller you make $\Delta t$, the more accurate your result will be; and in the limit that $\Delta t\to 0$, your result becomes exact (taking this limit is exactly the same thing as taking the derivative $\frac{d\textbf{P}}{dt}$, so this should be no surprise to you).