Question

The 8 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 0.4 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 8 kg block. Find the speed of the 8 kg block just before it collides with the 2 kg block. Answer in units of m/s.

Answer 1

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J$

As the energy in the system is conserved we have

$\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s$

The speed of the 8 kg block just before collision is 3.258 m/s