Answer :
Answer:
4.04m
Explanation:
First you calculate the velocity of the block when it leaves the spring. You calculate this velocity by taking into account that the potential energy of the spring equals the kinetic energy of the block, that is:
[tex]U=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(10000N/m)(0.08m)^2}{12kg}}=2.3\frac{m}{s}[/tex]
To find the distance in which the block stops you use the following expression (net work done by the friction force is equal to the difference in the kinetic energy of the block):
[tex]W_{T}=\Delta K\\\\F_f d=\frac{1}{2}m[v^2-v_o^2]\\\\d=\frac{mv^2}{2F_f}[/tex]
where Ff is the friction force. By replacing the values of the parameters you obtain:
[tex]d=\frac{(12kg)(2.3m/s)^2}{2(8N)}=3.96m[/tex]
hence, the distance to the original position is 3.96m+0.08m=4.04m