Question

A 5.17 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is fixed. The spring is compressed 0.102 m from equilibrium and is then released. The speed of the block is 1.30 m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which $\mu$k = 0.270. Determine the speed of the block at the equilibrium position of the spring.

Answer 1

Answer:

v=0.816 m/s

Explanation:

The force of the spring and the motion of the block are in equilibrium so without any force of friction the motion is

$E_{s}=E_{k}$

$\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2$

First determinate the constant of the spring that produce the kinetic energy of the bloc

$k=\frac{m*v^2}{d_{s}^2}$

$k=\frac{5.17kg*(1.30\frac{m}{s})^2}{0.102^2m}$

$k=839.8 \frac{kg}{s^2}$

Now the motion with the force of friction in the kinetic

$E_{s}=E_{k}-W_{k}$

$\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2-u*m*g$

Resolve to v

$v^2=\frac{(k*d_{s}^2)+(2*g*u)}{m}$

$v=\sqrt{\frac{839.8*(0.102m)^2-2*9.8*0.270}{5.17}}$

$v=0.81 \frac{m}{s}$