Answer:
The box slides 0.58 m before stopping .
Explanation:
Given :
Spring constant , k = 100 N/m .
Compression in the spring , x = 20 cm .
Mass of box , m = 2.3 kg .
Coefficient of kinetic friction of the box on the surface is , [tex]\mu=0.15[/tex] .
Therefore , conserving energy of the system :
[tex]K.E_i+U_i=K.E_f+U_f\\\\\dfrac{kx^2}{2}+0=0+\mu( mgx)[/tex]
Putting value in above equation we get :
[tex]\dfrac{100\times 0.2^2}{2}+0=0+0.15\times ( 2.3\times 10\times x)\\\\2=x\times 3.45\\x=0.58\ m[/tex]
Therefore , the box slides 0.58 m before stopping .
