Answer to Question #221956 in Physics for caleb

Question

An object that follows Hooke’s Law is hanging 2 m above the ground with a mass of 1.5 kg and a spring constant of 1.2 N/m. If the object is at rest when stretched 0.25 m, what is its total mechanical energy?

Answer 1

The total mechanical energy consists (in this case) of gravitational and elastic potential energies. The gravitational one is given as follows:

$$E_g = mgh$$

where $m = 1.5kg$ is the mass of the object, $h = 2m$ is the height above the ground, $g = 9.81m/s^2$ is the gravitational acceleration.

The elastic one:

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

where $k = 1.2N/m$ is the spring constant, $x = 0.25m$ is the elongatin of the spring.

Thus, obtain:

$$E = E_g + E_e = 1.5\cdot 2\cdot 9.81 + \dfrac{1.2\cdot 0.25^2}{2} \approx 29J$$

Answer. 29 J.