Question

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 5.5 cm has an energy of 28 J. If the block is replaced by one whose mass is twice the mass of the original block and the amplitude of the motion is again 5.5 cm, what is the energy of the system?

Answer 1

Answer:

The energy remains the same as  28J

Explanation:

The spring's total energy

$E_k=\frac{1}{2}*K*x^2$

Since k and A are not changed  

$K=\frac{2*E_k}{x^2}$

$K=\frac{2*28J}{0.055m^2}$

$K=18512.3 N/m$

the energy remains the same as  28J

$E_k=\frac{1}{2}*18512.3 N/m*0.055m^2$

$28J=\frac{56J}{2}$

$28J=28J$