Forces of 10.3N north, 19.5N east, and 15.2N south are simultaneously applied to a 4.06kg mass as it rests on an air table. What is the magnitude of its acceleration?

Answer 1

Fnet = F1 + F2+ . . .

Fnet =(10.3 N [North])+(19.5 N [South]) - 15.2 N [South] = 14.6 N [North]

Fnet = ma

a = Fnet/m

14.6 N [North]/(4.06 kg)

a = 3.60 m/s/s [North]

Answer 2

F = 10.3i + 19.5 - 15.2i = 19.5 - 4.9i.

X = 19.5, Y = -4.9.

F = Sqrt(X^2 + Y^2)=Sqrt(19.5^2+4.9^2) =
20.1 N.

a = F/M = 20.1/4.06 = 4.95 m/s^2.

Note: The direction(angle) is not required.