Question

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 9.3 N, acting 64° north of west. What is the magnitude of the body's acceleration?

Answer 1

Now , in order to get the net force we will have to calculate the vector sum ofr the two forces.

\(\sf \sqrt{9^2+9.3^2+2\times 9\times 9.3 \times cos135°}\approx 7\)

Answer 2

Acceleration = 7/3=2.33\(\sf m/s^2\)