A(n) 14 g bullet is fired into a(n) 121 g block of wood at rest on a horizontal surface andstays inside. After impact, the block slides 8.3 m...

The kinetic energy og the bullet  dragged the the block
along with the bullet  for a distance of 8.3meter.

The work
done here could be estimated by the frictional force F for a distance of 8.3 meter, 
which is F*s .

The frictional force = weight force of the
bolck with bullet*coefficient of friction* distance.

=
(14gram+12gram)(g m/s^2)(0.7) (8.3 m). Convert  everything in MKS
system.

= (0.014kg
+0.121kg)(9.8m/s^2)(0.7)(8.3m)

= (0.135)(9.8)(0.7)(8.3)
J

= 7.68663 J

We presume the
kinetic energy of the bullet has been converted into  7.68663 j of
energy.

So the KE of bullet is (1/2)mv^2, where m = bullets
mass = 14g = 0.014kg. And v is the velocity of the bullet before hit, to be
determined.

(1/2)(0.014)v^2 =
7.68663.

v^2 = 2*7.68663/0.014 =
1098.09)

v = sqrt(1098.09) = 33.14m/sec
approximately.

So the velocity of the bullet  slightly more
tham 33.14 m/s as a part of the energy has to be  converted into heat , noise  and
penetrating into the block etc.

=
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