Answer to Question #154557 in Physics for buddy

Question

A spring attached to the wall is connected to a block on the other end. The spring is stretched 47 cm away from the wall and then let go. The spring has a spring constant of 208 N/m. If the net force on the spring is 111 N 28° below the horizontal, and assuming the mass of the spring is negligible, what is the mass of the block?

Answer 1

First, calculate how much force the spring exerts on the wall with Hooke's law:

$$F=kx.$$

Second, remember that we can represent the net force as a vector sum of two forces by Pythagorean theorem:

$$F_\text{net}=\sqrt{F_h^2+F_v^2}.$$

In the last equation, the horizontal force is the force by Hooke's law, the vertical force is... the force of gravity!

$$F_\text{net}=\sqrt{(kx)^2+(mg)^2},\\\space\\ m=\frac{\sqrt{F^2_\text{net}-(kx)^2}}{g},\\\space\\ m=\frac{\sqrt{111^2-(208\cdot0.47)^2}}{9.8} =5.36\text{ kg}.$$