Recent content by Jose Miranda

ok, so an answer for this can be given as vectors? (22.60ihat N -14.6 jhat N) ?

The problem doesn't mention an incline i just assumed it was on an incline, because of its final velocity.

What stumps me is the how can i do this if there is no angle.

I believe the a I got the first time is wrong. a = (V-Vi)/t, so a = (6.1ihat - 3.95 jhat) m/s The problem is on on incline starting at 0,0 and having an angle below the x-line. So the force on the x-axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N 22.6 +14.6 =...

Homework Statement A 3.7 kg mass has a constant acceleration from an initial velocity of 2.33 ihat m/s to a final velocity of (18.9 ihat - 10.7 jhat) m/s in 2.71 s. Two forces act on the mass: gravity and an unknown applied force. Determine the acceleration and unknown force.[/B] Homework...