Question

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.

Answer 1

Answer:

Explanation:

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.

$K_1+U_{e1}+W_f=K_2+U_{e2}$

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

the friction force is given by:

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

The work is given by:

$W_f=F_f*d*cos(\theta)$

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

$W_f=9.9N*(0.0120m)*cos(180)\\W_f=-0.12J$

applying the theorem:

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

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