A1.7-kg block of wood rests on a rough surface. a 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself. the bullet-block system slides forward 2.4 m before coming to rest.

What is the coefficient of kinetic friction between the block and the surface?  Express your answer using two significant figures.

0.39

Explanation:

Given information

m1=1.7 Kg

m2=0.011 Kg

v2=670 m/s

d=2.4 m

m2v2=(m1+m2)v hence and also

The deceleration due to friction is given by

therefore

Therefore,

ill do it for 100, also this isnt 80 its 40

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