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
[tex]U = \dfrac{1}{2}kx^2[/tex]
[tex]U = \dfrac{1}{2}\times 350 \times 0.5^2[/tex]
U = 43.75 J
using conservation of energy
KE = U - f.d
where f is the frictional force acting
[tex]\dfrac{1}{2}mv^2 = 43.75- \mu m g d[/tex]
[tex]\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5[/tex]
[tex] v= \sqrt{11.643}[/tex]
v =3.41 m/s
