How to Calculate Net Force Magnitude at Specific Time?

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danielatha4
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Homework Statement


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.

Homework Equations


Our teacher has been making us use the position update formula and the momentum principle:
[tex]\Delta[/tex]r=Vavg[tex]\Delta[/tex]t
[tex]\Delta[/tex]P=Fnet[tex]\Delta[/tex]t

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
 
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danielatha4 said:

Homework Statement


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.


Homework Equations


Our teacher has been making us use the position update formula and the momentum principle:
[tex]\Delta[/tex]r=Vavg[tex]\Delta[/tex]t
[tex]\Delta[/tex]P=Fnet[tex]\Delta[/tex]t



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

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.
 
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?
 
danielatha4 said:
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?

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