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5 March, 15:56

A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equilibrium length. A 100 g hard plastic ball is then loaded into the tube. If the ball is shot directly up and reaches a height of 2 m above the top of the tube, what is the spring constant, k? Ignore air resistance.

A) 98 N/m

B) 20 N/m

C) 12 N/m

D) 25 N/m

E) 390 N/m

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  1. 5 March, 18:54
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    Answer: The spring constant is K=392.4N/m

    Explanation:

    According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

    The force F=ke

    Where k=spring constant

    e = Extention produced

    h=2m

    Given that

    e=20cm to meter 20/100 = 0.2m

    m=100g to kg m=100/1000 = 0.1kg

    But F=mg

    Ignoring air resistance

    assuming g=9.81m/s²

    Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

    E=1/2ke²=mgh

    Substituting our values to find k

    First we make k subject of formula

    k=2mgh/e²

    k=2*0.1*9.81*2/0.1²

    K=3.921/0.01

    K=392.4N/m
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