Finding magnitude of net force

[danielatha4]

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:
$$\Delta$$r=V_avg$$\Delta$$t
$$\Delta$$P=F_net$$\Delta$$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

 

[pgardn]

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:
$$\Delta$$r=V_avg$$\Delta$$t
$$\Delta$$P=F_net$$\Delta$$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.

 

[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?

 

[gabbagabbahey]

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 $t=t_0$ 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).

 

[paranormal]

.

To find the magnitude of the net force, we can use the momentum principle:

ΔP = FnetΔt

where ΔP is the change in momentum and Δt is the change in time.

First, we need to find the momentum at t = 2.10 seconds. We can use the position update formula to find the position at t = 2.10 seconds, and then use the velocity equations to find the velocity at that point.

At t = 2.10 seconds:
x = (5.40)(2.10)^2 - 1.30 = 23.31 meters
y = (2.75)(2.10)^3 + 1.80 = 13.42 meters

Using the velocity equations:
vx = 10.8(2.10) = 22.68 m/s
vy = 8.25(2.10)^2 = 36.79 m/s

Now we can calculate the initial and final momenta:
Pinitial = (2.60)(0) + (2.60)(22.68) = 58.87 kg*m/s
Pfinal = (2.60)(23.31) + (2.60)(36.79) = 149.40 kg*m/s

Substituting these values into the momentum principle equation:
Fnet = (Pfinal - Pinitial)/Δt
Fnet = (149.40 - 58.87)/(2.10 - 0)
Fnet = 45.46 N

Therefore, the magnitude of the net force acting on the mass at t = 2.10 seconds is 45.46 N.