# 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 before coming the rest. find the speed of the bullet...

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 before coming the rest. find the speed of the bullet before impact. g = 9.8. If the coefficient of friction between the surface and the block is 0.7.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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william1941 | College Teacher | (Level 3) Valedictorian

Posted on

The mass of the block is 121g or .121 kg. As the bullet is lodged in the block the total mass is 121+14 = 135 g = .135 kg. The frictional force that makes the block come to a stop is normal force* coefficient of friction = .135*9.8*0.7 = 0.9261 N

As the block comes to rest after sliding for 8.3 meters the energy it was given by the bullet is 0.135*9.8*0.7*8.3 =7.68663 Nm

Now this energy is provided by the bullet. So the energy in the bullet was equal to 0.5*mass*velocity^2 = 0.5*14*v^2.

0.5*.014* v^2 = 0.135*9.8*0.7*8.3 =7.68663

=> .007* v^2 = 7.68663

=> v^2 = 7.68663/.007

=> v = sqrt [ 1098.09 ]

=> v = 33.13 m/s

So the speed of the bullet before impact was 33.13 m/s.