To solve this problem we will apply the concepts related to energy conservation. In this case the kinetic energy will be equivalent to the potential energy of the system. Mathematically said expression can be determined in the form:
[tex]\frac{1}{2} mv_0^2 = \frac{1}{2} kx^2[/tex]
Here,
m = Mass
[tex]v_0[/tex] = Initial velocity
k = Spring constant
x = Displacement
Our values are given as,
[tex]m = 1.2 kg[/tex]
[tex]v_0 = 6 m/s[/tex]
[tex]x = 0.10 m[/tex]
We use the previous equation to find the value of the spring constant,
[tex]\frac{1}{2} mv_0^2 = \frac{1}{2} kx^2[/tex]
[tex]k = \frac{mv_0^2}{x^2}[/tex]
Replacing,
[tex]k = \frac{(1.2)(6)^2}{(0.1)^2}[/tex]
[tex]k = 4320N/m[/tex]
The total energy and the spring-mass system will be
[tex]E = \frac{1}{2} kA^2[/tex]
Here
A = Amplitude
Replacing we have that
[tex]E = \frac{1}{2} (4320)(0.1)^2[/tex]
[tex]E = 21.6J[/tex]
Therefore the total energy of the block-spring system is 21.6J
