A weight suspended from a spring vibrates vertically with an amplitude of 4 cm.

A weight suspended from a spring vibrates vertically with an amplitude of 4 cm. Determine the total vibration energy of the weight if the coefficient of spring stiffness is 1 μN.

A = 4 cm = 0.04 m.

k = 1 μN = 0.000001 N / m.

E -?

The total mechanical energy of the body, E, consists of the kinetic Ek and the potential En energy of the body: E = Ek + En.

The potential energy En is determined by the formula: En = k * x ^ 2/2, where k is the stiffness of the spring, x is the deviation from the equilibrium position.

The kinetic energy Ek is determined by the formula: Ek = m * V ^ 2/2, where m is the mass of the load, V is the speed of the body.

Amplitude A is the maximum deviation of the body from the equilibrium position. In this position, the body stops, therefore, the total mechanical energy in this position consists only of potential energy.

E = Ep.

E = k * A ^ 2/2.

E = 0.000001 N / m * (0.04 m) ^ 2/2 = 8 * 10 ^ -10 J.

Answer: the total mechanical energy of the weight is E = 8 * 10 ^ -10 J