A 1.01 kg mass is attached to a spring of force constant 10.1 N/cm and placed on a frictionless surface. By how much will the spring stretch

Question

A 1.01 kg mass is attached to a spring of force constant 10.1 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.515 m at a rate of 1.84 revolutions per second

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Vân Khánh 3 years 2021-07-27T11:43:02+00:00 1 Answers 15 views 0

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  1. Nick
    0
    2021-07-27T11:44:12+00:00
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    Answer:

    6.88 cm

    Explanation:

    Given that:

    mass m of the object  = 1.0 kg

    Force constant of the spring (k) = 10.1 N/cm = 10.1 × (100) N/m

    = 1010 N/m

    Radius of the circular path R = 0.515 m

    Since 1 revolution = 2 π radian

    angular speed of the object ω  = 1.84 rev/s = 1.84 × ( 2 π) rad/s

    = 11.56 rad/s

    Amount of the spring that stretches = X???

    However, the amount of the force the spring exerts on the object due to the stretching is directly equal to the centripetal force required by the object for the circular motion:

    SO;

    K(X) = m \omega^2 R\\1010*(X) = 1.01 *(11.56)^2 *0.515 \\(X) = \frac{1.01*(11.56)^2*0.515}{1010} \\(X) = 0.0688

    X = 0.0688 m

    Hence, the amount of the spring stretches = 6.88 cm

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