Question

The 6 kg block is then released and accelerates

Answer 1

Answer:

5.709 m/s

Explanation:

Assuming compression of spring as 0.6m and spring constant of 602 N/m

The initial energy stored in the spring will be given by $\frac {kx^{2}}{2}$

By substitution, the initial energy is \$frac {602\times0.6^{2}}{2}=108.36 J$

Kinetic energy of the 6 Kg block is given by subtracting energy lost due to friction from the initial stored energy hence

$\frac {mv^{2}}{2}=108.36-\mu_k mgx$ where m is mass, v is speed, g is acceleration due to gravity, x is the spring compression and $\mu_x$ is the coefficient of friction

Making v the subject then  

$v^{2}=\frac {2(108.36-\mu mgx)}{m}$

$v=\sqrt{\frac {2(108.36-\mu_x mgx)}{m}}= \sqrt {\frac {2(108.36-(0.3\times 6\times 9.8\times 0.6))}{6}}\approx 5.709 m/s$