Answer to Question #309956 in Physics for ali

Question

Find the change in kinetic energy by an elastic spring with spring constant of 100 N/cm, as it goes back to its equilibrium position, from a length of 10 cm to 15 cm.

a. -250 J

b. -1250 J

c. 0

d. -2500 J

Answer 1

The potential energy stored in the compressed string is given as follows:

$$E_p = \dfrac{kx^2}{2}$$

where $x = 15cm-10cm = 5cm = 0.05m$ is the contraction, $k = 100N/cm = 10^4N/m$ is the spring constant.

As the spring returns in the equilibrium position, all this potential energy converts to the kinetic energy. Thus, the change in kinetic energy will be:

$$\Delta K = \dfrac{kx^2}{2} = \dfrac{10^4N/m\cdot (0.05m)^2}{2} = 12.5J$$

Answer. 12.5J.