Charles, the 75 kg trampoline artist, lands on a trampoline with a speed of 9.0 m/s.
If the trampoline behaves like a spring with a spring constant of 52,000 N/m, what maximum distance will Charles push down the trampoline before bouncing back up? (Hint: at maximum compression, Charles is not moving.)

Answers

Answer 1

Answer:

The maximum distance Charles will push down the trampoline ≈ 0.342 m

Explanation:

The given parameters are;

The mass of the trampoline artist, m = 75 kg

The speed with which the trampoline artist lands, v = 9.0 m/s

The value of the spring constant of the trampoline, k = 52,000 N/m

Let x represents the maximum distance Charles will push down the trampoline

Therefore, we have;

Kinetic energy = 1/2·m·v²

The kinetic energy with which the trampoline artist lands  = 1/2 × 75 × 9.0² = 3037.5

The kinetic energy with which the trampoline artist lands  =  3037.5 J

The potential energy stored in a spring = 1/2·k·x² = The kinetic energy with which the trampoline artist lands

∴ 1/2 × 52,000 × x² = 3037.5

∴ x = √(3037.5/(1/2 × 52,000)) ≈ 0.342

The maximum distance Charles will push down the trampoline = x ≈ 0.342 m


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Learn more about the orbital time period here:

https://brainly.com/question/14494804?referrer=searchResults

The attached picture shows the derivation of the formula for orbital speed.

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