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Posted by on Wednesday, October 26, 2011 at 11:09pm.

A 13.2 gm bullet is fired into a 12.0 kg wood block that is at rest on a wood table. The block, with the embedded bullet, slides 5.00 cm across the table. Assuming that the local acceleration due to gravity is 9.80m/s2 and that the coefficient of kinetic friction between the block and the table is 0.190, what was the speed of the bullet (in m/s)?

  • Physics help please - , Wednesday, October 26, 2011 at 11:51pm

    Use the distance that the bullet and block travel after impact (X = 0.0500 m)to get the initial kinetic energy of bullet and block together.

    (Friction force)*(distance)
    = (1/2)(M+m) Vo^2
    M + m = 12.0132 kg
    Friction force = (0.190)(M+m) g
    = 22.37 N

    Solve for Vo

    Once you have Vo, compute the initial momentum (M+m) Vo

    That equals the initial momentum of the bullet. Divide that by bullet mass m for the bullet's velocity.

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