Question

The 6 kg block is then released and accelerates to the right, toward the 4 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 4 kg block. Answer in units of m/s.

Answer 1

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 350 \times 0.5^2$

   U = 43.75 J

using conservation of energy

 KE = U - f.d

where f is the frictional force acting

$\dfrac{1}{2}mv^2 = 43.75- \mu m g d$

$\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5$

$v= \sqrt{11.643}$

       v =3.41 m/s