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The 6 kg block is then released and accelerates
to the right, toward the 2 kg block. The
surface is rough and the coefficient of friction
between each block and the surface is 0.3 . The
two blocks collide, stick together, and move
to the right. Remember that the spring is not
attached to the 6 kg block.
Find the speed of the 6 kg block just before
it collides with the 2 kg block.
Answer in units of m/s

Answer :

opudodennis

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 [tex]\frac {kx^{2}}{2}[/tex]

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

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

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

Making v the subject then  

[tex]v^{2}=\frac {2(108.36-\mu mgx)}{m}[/tex]

[tex]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[/tex]

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