Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.2 m and is released with a

Question

Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.2 m and is released with a speed of 12 m/s .

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Eirian 4 years 2021-08-24T23:32:19+00:00 1 Answers 424 views 0

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  1. kimcuc
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    2021-08-24T23:33:58+00:00
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    Answer:

    229.11 N

    Explanation:

    Mass of the shot (m) = 7.0 kg

    Distance moved by the shot (S) = 2.2 m

    Initial velocity of the shot (u) = 0 m/s (assume)

    Final velocity of the shot (v) = 12 m/s

    First we will determine the acceleration of the shot using the following equation of motion.

    v^2=u^2+2aS\\\\(12)^2=0+2a(2.2)\\\\144=4.4a\\\\a=\frac{144}{4.4}=32.73\ m/s^2

    So, the acceleration of the shot is 32.73 m/s².

    Now, using Newton’s second law, the force acting on the shot is equal to the product of mass and acceleration. So,

    Force = Mass × Acceleration

    F=ma\\\\F=7\ kg \times 32.73\ m/s^2\\\\F=229.11\ N

    Therefore, the average force exerted by the shot-putter is 229.11 N.

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