Answer to Question #224944 in Physics for alyssa

Question

The length of the spring becomes 29cm when a load of 6kg is hung on it. If the elastic potential energy stored in the spring is 2.7J, calculate the original length of the spring?

Answer 1

Express the spring constant from energy:

$$E=\frac12 k\Delta x^2,\\\space\\ E=\frac12 k \Delta x^2,\\\space\\ k=\frac{2E}{\Delta x^2}.$$

By Hooke's law:

$$mg=k\Delta x,\\\space\\ k=\frac{mg}{\Delta x}.$$

Substitute:

$$\frac{mg}{\Delta x}=\frac{2E}{\Delta x^2},\\\space\\ \Delta x=\frac{2E}{mg}=9.2\text{ cm}.$$

What is $\Delta x$?

$$\Delta x=x_f-x_i,\\\space\\ x_i=x_f-\Delta x=29-9.2=19.8\text{ cm}.$$