Question

A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmonic oscillations with an amplitude of 7.50 cm. What is the total mechanical energy of the system

Answer 1

Answer:

E = 0.01 J

Explanation:

Given that,

The mass of the cart, m = 0.15 kg

The force constant of the spring, k = 3.58 N/m

The amplitude of the oscillations, A = 7.5 cm = 0.075 m

We need to find the total mechanical energy of the system. It can be given by the formula as follows :

$E=\dfrac{1}{2}kA^2$

Put all the values,

$E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J$

So, the value of total mechanical energy is equal to 0.01 J.