One end of a vertical 360 N/m spring is attached to the ceiling and the other end is at- tached to a 20 kg mass. The mass is released from rest when the spring is at equilibrium length. The mass falls 1.2 m before again coming to rest, how much mechanical energy is lost during this fall

259.2 Joule

Explanation:

Spring constant, K = 360 N/m

mass, m = 20 kg

Amplitude, A = 1.2 m

According to the equilibrium of force

The mechanical energy lost is given by

U = 0.5 KA²

U = 0.5 x 360 x 1.2 x 1.2

U = 259.2 Joule

learn it on the intertant so you can solve

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6