Need some help with an AP Physics problem.?

Question

A particle of mass 0.4kg is subjected to two forces F1=2Ni-4Nj and F2= -2.6Ni+5Nj. If the particle starts from rest at the origin at t=0 find its position and velocity at t=1.6 sec.

by position i assume they mean distance.

Answer 1

[Edit -- two things. #1 Oops, I made a mistake. When I worked this, I used a mass of 4 kg instead of 0.4. Now I'll go back through my answer and make the correction.

[#2 There are two different methods here. Look at both of them. The way is for this problem. I think you'll see that.]

Snow White has an answer, but she doesn't say how she got it.

Let's start by getting the net force F = F1 + F2 = -0.6Ni + Nj with magnitude sqrt(1.36)N = 1.166 N and acceleration F/m = 2.915476 m/s^2.

Now we have two ways to proceed. This is linear motion with constant acceleration, and we know the direction ... tan theta = -1 / 0.6 = -5/3, so the angle theta in the second quadrant is 59 degrees north of west.

After 1.6 seconds, the velocity is acceleration x time = 2.915476 x 1.6 = 4.66476 m/s in the direction indicated above.

The displacement s after 1.6 seconds is s = vt/2 or s = 1/2 at^2 = 4.66476 x 1.6 / 2 = 3.7318 m.

To get the position, we could use sines and cosines, but notice that tan theta above is -5/3, so the hypoteneuse of that triangle is sqrt(25+9) = sqrt(34). Using that geometry, the position after 1.6 seconds is

x = -3/sqrt(34) x 3.7318 = -1.92 m

y = 5/sqrt(34) x 3.7318 = 3.2 m

That's your answer: (-1.92, 3.2) with velocity 4.66476 m/s and direction 59 degrees north of west.


There's another way to do it that may be better, especially when you're using vectors. Do it by components, and keep the components separate.

The net force is F = F1 + F2 = -0.6Ni + Nj.

The constant acceleration a = F/m = -1.5i + 2.5j m/s^2

The velocity after 1.6 seconds is v = at = -2.4i + 4j m/s

The position after 1.6 seconds is s = vt/2 = -1.92i + 3.2j

and this is the same answer we got by doing it the other way. (You can, of course, get all the magnitudes and angles.)

Looks to me that it's easier doing it by components.

Hope this is the same answer Snow White got. This way, at least, you have two ways of doing it.

Answer 2

let's determine the direction and magnitude of the force"

F_total= 2Ni-4Nj -2.6Ni+5Nj= - 0.6Ni + 1Nj

magnitude = (0.6^2 + 1^2)^0.5 = 1.166 N

angle ====> tan(A) = 1/(-0.6), A= 59 degrees N of W

now we can find acceleration

F=mxa, a=1.166 N / 0.4 kg = 2.915 m/s2

next is velocity

a = (V_final - V_initial)/(delta t), V_initial = 0, t=1.6 sec

V_final=2.915*1.6=4.66 m/s,

total distance=V_initial*t + 0.5*a*t^2 = 0 + 0.5* 2.915*1.6^2
=3.7312 m

x position = -3.7312cos59 = -1.92 m
y position = 3.7312sin59= 3.2 m

Answer 3

the velocity is 4.665m/s
3.74m, (-1.93, 3.21)
59 degrees north of west

i believe that's rite

Answer 4

wow sounds complex..did you by any chance take ap chemistry? thats what i'm in now.,.its tough stuff