Inelastic collision & friction

[Hawraa]

the question is :
a 1 kg block at rest on a horizontal surface is connected to an unstreched spring with a spring constant k=1000 N/m . A 2 kg block whose speed is 6 m/s collides with 1kg block. if the two blocks stick together and move on a rough surface after collision: find
1- the final velocity just after collision in ( m/s ) ?

i found this using momentum equation
m1v1=m2v2
(2)(6) = (2+1)v2
v2( which is the common velocity that the 2 masses are going to move with )= 12/3=4m/s

now the second part
find the coefficient of the kinetic friction of the rough surface , if the spring is compressed by maximum distance of x=0.2 m ?

i tried to solve it using this equation but finally i get a coefficient which is bigger than 1 !

delta K + delta U = work done by friction
{(1/2)(3)(4)square - (1/2)(2)(6)square } + { (1/2)(1000)(0.2)square - (1/2)(1000)(zero)square} = work done by friction

wf = 8
- kinetic friction coefficient . Normal force . d = 8
- kinetic friction coefficient . (3)(10) . (0.2) = 8
i get this answer kinetic friction coefficient = -1.3
canu pleez tell me where did i go wrong in this ?

 

[Doc Al]

delta K + delta U = work done by friction
{(1/2)(3)(4)square - (1/2)(2)(6)square } + { (1/2)(1000)(0.2)square - (1/2)(1000)(zero)square} = work done by friction

Not sure what you're doing here. The starting point is immediately after the collision--the 6 m/s speed is no longer relevant. Compare the initial and final energy of the 2-mass system.

 

[Hawraa]

so i consider the final velocity is zero ?? as the spring compresses to the maximum distance ?

 

[Doc Al]

so i consider the final velocity is zero ?? as the spring compresses to the maximum distance ?

Sure. As long as it's still moving, it hasn't reached full compression. Right?

 

[Hawraa]

thank u i got it 0.6 i think it's right :D