Question

Find the mechanical energy of a block-spring system having a spring constant of 1.3 N/cm and an oscillation amplitude of 2.4 cm

Answer 1

mechanical energy of block-spring system = 1/2 * k * x^2
k = 1.3 N/cm = 130 N/m
x = 2.4 cm = 0.024 m
hence

energy = 0.5 * 130 * 0.024^2 = 0.03744 Joules

Answer 2

as the amplitude is the extreme position total energy is of spring block is used to stretch the spring
M.E = 1/2(130)(2.4/100)^2 =0.03744 J

Answer 3

E=K+P
E=1/2(mv^2 + Kx^2)
E=1/2KA^2
E=0.5(1.3)(2.4^2)= 3.744 N cm