Question

a 0.50-kg block attached to an ideal spring with a spring constant of 80 n/m oxalates and horizontal frictionless surface the total mechanical energy point. 4 the greatest extension of the spring from its equilibrium length is

Answer 1

The greatest extension in the spring will happen when velocity of block reduces to 0 m/s. at this moment all the kinetic energy of block have been converted to potential energy of spring. and block has no kinetic energy.
So, PE of spring = total mechanical. energy
i.e. ,
$(1 \div 2)k\times {x}^{2} = 0.4 \\ x = 10 \: cm$
assuming in question that total mechanical energy is 0.4 J.