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

E = 0.01 J

Step-by-step 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=(1)/(2)kA^2`

Put all the values,

`E=(1)/(2)* 3.58* (0.075)^2\\\\=0.01\ J`

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