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A 2-oz bullet is fired horizontally with a velocity v1 = 1800 ft/sec into the 6-kg block of soft wood initially at rest on the horizontal surface. The bullet emerges from the block with the velocity v2 = 1200 ft/sec, and the block is observed to slide a distance of 8 ft before coming to rest. Determine the coefficient of kinetic friction μk between the block and the supporting surface.

Respuesta :

Answer:0.0623

Explanation:

 Given

Initial velocity of bullet=1800 ft/sec

Final velocity of bullet=1200 ft/sec

mass of bullet =2 oz=56.699 gm

Now conserving momentum

[tex]m_{bullet}u_i=m_{bullet}u_f+m_{block}v[/tex]

[tex]56.699\times 1800=56.699\times 1200+6000\times v[/tex]

v=5.669 ft/sec[/tex]

Now applying equation of motion to get Coefficient of friction

[tex]v_f^2-u_i^2=2as[/tex]

where [tex]a=\mu _k\times g[/tex]

here [tex]u_i=5.669 [/tex]

and [tex]v_f=0[/tex]

[tex]5.669^2=2\times \mu _k\times g\times 8[/tex]

here [tex]g=32.2 ft/s^2[/tex]

[tex]\mu _k=0.0623[/tex]

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