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.

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;

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

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.