A 2.60kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0310m . The spring has force constant 810N/m . The coefficient of kinetic friction between the floor and the block is 0.39 . The block and spring are released from rest and the block slides along the floor.

A-What is the speed of the block when it has moved a distance of 0.0120m from its initial position? (At this point the spring is compressed 0.0190m .)

Express your answer with the appropriate units.

In order to solve this problem we need to apply the Energy Conservation Theorem, The motion occurred on the ground so the potential gravitational energy is zero.

[tex]K_1+U_{e1}+W_f=K_2+U_{e2}[/tex]

We need to calculate the work done by the friction force.

the friction force is given by:

[tex]F_f=\µ*F_N\\F_f=\µ*m*g\\F_f=0.39*2.60kg*9.8m/s^2\\F_f=9.9N[/tex]

The work is given by:

[tex]W_f=F_f*d*cos(\theta)[/tex]

The angle of the force is 180 degrees because it is opposite to the motion.

[tex]W_f=9.9N*(0.0120m)*cos(180)\\W_f=-0.12J[/tex]

applying the theorem:

[tex](0)+\frac{1}{2}*810N/m*(0.0310m)^2-0.12J=\frac{1}{2}*2.60*v^2+\frac{1}{2}*810N/m*(0.0190m)^2[/tex]

[tex]0.269J=\frac{1}{2}*2.60*v^2+0.146J\\v=\sqrt{\frac{2*(0.269J-0.146J)}{2.60kg}}\\\\v=0.308m/s[/tex]