Question

The weight of the load attached to the spring is 50 g. The spring hardness is 1000 n / m. The load was shifted 10 cm from the equilibrium position and released. What speed will the cargo have when passing through the equilibrium state.

Answer 1

According to work-energy theorem, the speed will be

$$\Delta W=\frac{mv^2}{2},\\\space\\ \frac{k\Delta x^2}{2}=\frac{mv^2}{2},\\\space\\ v=\Delta x\sqrt{\frac{k}{m}}=0.1\sqrt{\frac{1000}{0.05}}=14\text{ m/s}.$$