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physics

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A 20 g bullet moving horizontally with a speed v hits a 2-kg block of wood resting on a table. After hitting the block the bullet is embedded in the block of wood and the block and the bullet moves a distance of 40 cm before it stops. If the coefficient of friction between the block and the table is 0.2 find the initial speed v of the block.

  • physics -

    mass of block with bullet = 2+.020 =2.02 kg
    weight of block with bullet = 2.02*9.81 = 19.8 N
    friction force = 19.8*.2 = 3.96 N
    work done by friction = 3.96 * .40 = 1.59 J
    that is kinetic energy of block with bullet
    1.59 = (1/2)(2.02) v^2
    v = 1.25 m/s for block with bullet
    now conservation of momentum
    2.02 * 1.25 = .020 * v
    v = 127 m/s

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