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 len

Question

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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18 hours 2021-07-22T08:10:46+00:00 1 Answers 0 views 0

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    2021-07-22T08:12:26+00:00

    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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