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18 September, 14:21

A (n) 14 g bullet is fired into a (n) 121 g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 8.3 m before coming to rest. The acceleration of gravity is 9.8 m/s 2. If the coefficient of friction between the surface and the block is 0.7, find the speed of the bullet before impact. Answer in units of m/s

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  1. 18 September, 14:41
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    33.14 m/s

    Explanation:

    The mass of the block is 121g or. 121 kg. As the bullet is lodged in the block the total mass is 121+14 = 135 g or 0.135 kg.

    The frictional force that makes the block come to a stop is normal force * coefficient of friction = 0.135 * 9.8 * 0.7 = 0.9261 N

    As the block comes to rest after sliding for 8.3 meters the energy it was given by the bullet is

    0.135 * 9.8 * 0.7 * 8.3

    = 7.69 Nm

    Now this energy is provided the bullet. So the energy in the bullet was equal to

    1/2 * mv² = 0.5 * 14 * v².

    0.5 * 0.014 * v^2 = 0.135 * 9.8 * 0.7 * 8.3 = 7.69

    => 0.007 * v² = 7.69

    => v² = 7.69 / 0.007

    => v² = 1098.57

    => v = √1098.57

    => v = 33.14 m/s
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