contestada

A small block, with a mass of 0.05 kg compresses a spring with spring constant 350 N/m a distance of 4 cm. It is released from rest, then slides around the loop and up the incline before momentarily comes to rest at point A. The radius of the loop is 0.1 m.

Required:
Find the elastic potential energy.

Respuesta :

onyebuchinnaji

Answer:

The elastic potential energy of the spring is 0.28 J

Explanation:

Given;

mass of the block, m = 0.05 kg

spring constant, k = 350 N/m

extension of the spring, x = 4 cm = 0.04 m

The elastic potential energy of the spring is calculated as;

[tex]U_x = \frac{1}{2}kx^2\\\\U_x = \frac{1}{2} \times 350 \times (0.04)^2\\\\U_x = 0.28 \ J[/tex]

Therefore, the elastic potential energy of the spring is 0.28 J

Otras preguntas