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A child bounces in a harness suspended from a door frame by three elastic bands. (a) If each elastic band stretches 0.240 m while supporting a 6.55-kg child at rest, what is the force constant for each elastic band? Assume that each spring supports 1/3 of the child's weight. n/m (b) What is the time for one complete bounce of this child? s (c) What is the child's maximum velocity if the amplitude of her bounce is 0.240 m?

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Final answer:

The force constant for each elastic band can be calculated using Hooke's law, and it is approximately 267.96 N/m. The time for one complete bounce of the child is approximately 0.954 seconds. The child's maximum velocity during the bounce is approximately 0.634 m/s.

Step-by-step explanation:

The force constant of an elastic band can be calculated using Hooke's law, which states that the force exerted by a spring is directly proportional to its extension. In this case, each elastic band stretches 0.240 m while supporting 1/3 of the child's weight. So, the force constant for each elastic band can be calculated by dividing the weight of the child by the extension:

Force constant = (Child's weight / Extension)

For example, if the child weighs 6.55 kg and the extension is 0.240 m, the force constant for each elastic band would be:

Force constant = (6.55 kg × 9.8 m/s²) / 0.240 m = 267.96 N/m

The time for one complete bounce of the child can be calculated using the formula:

Time = 2π√(Mass / Force constant)

Using the given mass (6.55 kg) and the force constant (267.96 N/m), we can calculate the time:

Time = 2π√(6.55 kg / 267.96 N/m) ≈ 0.954 seconds

The child's maximum velocity during the bounce can be calculated using the formula for simple harmonic motion:

Maximum Velocity = Amplitude × √(Force constant / Mass)

With an amplitude of 0.240 m, a force constant of 267.96 N/m, and a mass of 6.55 kg, we can calculate the maximum velocity:

Maximum Velocity = 0.240 m × √(267.96 N/m / 6.55 kg) ≈ 0.634 m/s

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