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2 January, 11:01

A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tree's trunk. What is the initial kinetic energy of the bullet? 153.125 J Correct: Your answer is correct. How much work does the tree do in stopping the bullet?

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  1. 2 January, 11:09
    0
    Answer: Work done = 153.125Joules, Work done = 0.003Nm

    Explanation:

    Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

    Mathematically,

    K. E = 1/2MV²

    Where;

    M = mass of the body = 2.5g = 0.0025kg

    V = velocity of the body = 350m/s

    Substituting this values in the formula, we have;

    K. E = 1/2 * 0.0025*350²

    K. E = 153.125Joules

    Work done is the force applied to body to cause it to move through a distance.

    Work = Force * distance

    Force = ma = 0.0025 * 10

    Force = 0.025N

    Distance = 12cm = 0.12m

    Work = 0.025*0.12

    Work = 0.003Nm

    work done by the tree in stopping the bullet is 0.003N
  2. 2 January, 11:14
    0
    The work done by the tree in stopping the bullet is 0.00294 J

    Explanation:

    To determine how much work done by the tree in stopping the bullet, we first determine the force of the bullet.

    Force = mass x acceleration due to gravity

    Force = 0.0025 x 9.8 = 0.0245 N

    Work done = Force x distance

    = 0.0245 x 0.12

    = 0.00294 J

    Therefore, the work done by the tree in stopping the bullet is 0.00294 J
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