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IMPORTANT! A 50.0 g bullet is fired horizontally into a 4.50 kg wooden block that is initially at rest on a horizontal table. The initial speed of the bullet is 95.0 m/s. Upon impact, the block with the bullet slides across the table for a distance d before coming to a stop. Given that the coefficient of friction between the block and the table is 0.250, how far (d) did the block with the embedded bullet slide?

Respuesta :

Answer:

20.22 m

Explanation:

Bullet mass = .05 kg

Initial Kinetic Energy = 1/2 m v^2 =  1/2 (.05)(95^2 ) = 225.625 j

The work of friction must equal this KE

Normal force will be the wooden block PLUS the bullet times  g

 =  4.55 * 9.81

  force of friction is normal force * coefficient of friction

    =  4.55 * 9.81 * .250 N

  work of friction will be the force of friction x distance

       

 225.625  J  =  (4.55)(9.81)(.250) * d

   d = 20.22 m

 

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