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anonymous

  • one year ago

Tutoring;Mathematical Representation of a Spring

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  1. anonymous
    • one year ago
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    A 2.0 kg mass is fastened to a spring with force constant of 45 N/m. The mass is held motionless by a student at a position 15 cm below the spring's natural unstretched position. When the ball is released, what is the Fnet acting on it? What is the acceleration of the ball at that point?

  2. anonymous
    • one year ago
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    Here note that k constant is 45N/m as well as that from point of reference of the spring stretch is revealed to be 0.15m

  3. anonymous
    • one year ago
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    First of all we need to know the net force acting on the ball before we can calculate the acceleration. The net force acting on the spring can be calculated as 45N/m*0.15m=6.75N Given that the ball weighs 2.0kg, using the equation F=ma 6.75N=(2.0kg)(x) 6.75N/2.0kg=3.375m/s Therefore the resulting acceleration when the ball is stretched is 3.375m/s.

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