Question

A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. The upper end of the spring is depressed 0.10 m, and a 2.0-kg block is placed against the depressed spring. The system is then released from rest. What is the kinetic energy of the block at the instant it has traveled 0.10 m and the spring has returned to its uncompressed length?

Answer 1

Answer:

kinetic energy  is 2.02 J

Explanation:

Given data

k = 600 N/m

angle = 30°

spring = 0.10 m

mass = 2 kg

to find out

kinetic energy of the block

solution

we know initial energy that is

Energy 1 = 1/2 × kx²

put here k = 600 and x = 0.1

energy 1 = 1/2 × 600(0.1)²

energy  = 3

potential energy  PE = mgh

PE = 2 ×9.8 ×0.1sin30

PE = 0.98 J

so we know final energy is

E = KE of block + PE

so

kinetic energy  = 3 - 0.98

kinetic energy  = 2.02 J