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A 15.0 gram bullet strikes and becomes embedded in a 1.24 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.280, and the impact drives the block a distance of 11.0 m before it comes to rest, what was the muzzle speed of the bullet?

  • Physics - ,

    Vb = vel of bullet
    initial momentum = .015 Vb
    =final momentum = 1.255 Vi
    friction force = 1.255(.280)(9.81)

    initial KE = (1/2)(1.255) Vi^2
    work done = inital Ke = 1.255(.280)(9.81)(11)
    solve that for Vi
    Vb = (1,255/.015)Vi

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