Question

A horizontal spring with a 10000 N/m spring constant is compressed 0.08 m, and a 12-kg block is placed against it. When the block is released, the block shoots forward along a horizontal surface that exerts 8 N friction force on the block. How far from the original position does the block travel before coming to a stop

Answer 1

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:

$U=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(10000N/m)(0.08m)^2}{12kg}}=2.3\frac{m}{s}$

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):

$W_{T}=\Delta K\\\\F_f d=\frac{1}{2}m[v^2-v_o^2]\\\\d=\frac{mv^2}{2F_f}$

where Ff is the friction force. By replacing the values of the parameters you obtain:

$d=\frac{(12kg)(2.3m/s)^2}{2(8N)}=3.96m$

hence, the distance to the original position is 3.96m+0.08m=4.04m