Answer to Question #156551 in Physics for buddy

Question

A 2.2 kg block is attached to a spring that is attached to a vertical wall. The spring is stretched by 9 cm away from the wall and has a spring constant of 90 N/m. Assuming the mass of the spring is negligible, what is the magnitude and direction of the net force on the block? (Give the angle with respect to the horizontal, to the nearest degree.)

Answer 1

The magnitude of resultant force

$$F=\sqrt{(kx)^2+mg}\\ =\sqrt{(90*0.09)^2+(2.2*9.8)^2}=23\;\rm N$$

Angle

$$\theta=\tan^{-1}\frac{mg}{kx}=\tan^{-1}\frac{2.2*9.8}{90*0.09}=69^{\circ}$$