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Trampoline PE Grace has a mass of m. She drops a distance of h from a platform onto a trampoline which acts like a vertical spring with spring constant k. The maximum force experienced by Grace's legs (when the trampoline is at its lowest point) is n times larger than Grace's weight. Ignore air resistance. 7 m h k 25.2kg 0.896m ✓ (no answer) Correct Answer: 2.47 158 N m n Solution: This solution uses a reference frame with y = 0 at the initial level of the trampoline. Conservation of Energy with (initial=before drop) and (final=trampoline at lowest point, stretched down a distance x): PE¿ + KE¿ + Wnc =PEf+KEf mg(h)+0+0 = mg(−x) + ¾½k(x)² +0 (Eqn. 1) The variable x is not in the problem statement. It can be related to the variable n by applying Hooke's Law at the trampoline's lowest point: |Flegs| = kx = n(mg) x = (mg) n Plug into (Eqn. 1) and rearrange terms: (mg)2 n² - 2k (mg)2 k :) n n-mgh = 0 2k Clean up by multiplying by (mg)2 : n² – 2n − ( 2kh = 0 mg

Trampoline PE Grace has a mass of m. She drops a distance of h from a platform onto a trampoline which acts like a vertical spring with spring constant k. The maximum force experienced by Grace's legs (when the trampoline is at its lowest point) is n times larger than Grace's weight. Ignore air resistance. 7 m h k 25.2kg 0.896m ✓ (no answer) Correct Answer: 2.47 158 N m n Solution: This solution uses a reference frame with y = 0 at the initial level of the trampoline. Conservation of Energy with (initial=before drop) and (final=trampoline at lowest point, stretched down a distance x): PE¿ + KE¿ + Wnc =PEf+KEf mg(h)+0+0 = mg(−x) + ¾½k(x)² +0 (Eqn. 1) The variable x is not in the problem statement. It can be related to the variable n by applying Hooke's Law at the trampoline's lowest point: |Flegs| = kx = n(mg) x = (mg) n Plug into (Eqn. 1) and rearrange terms: (mg)2 n² - 2k (mg)2 k :) n n-mgh = 0 2k Clean up by multiplying by (mg)2 : n² – 2n − ( 2kh = 0 mg

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter7: Rotational Motion And Gravitation
Section: Chapter Questions
Problem 32P
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Trampoline PE Grace has a mass of m. She drops a distance of h from a platform onto a trampoline which acts like a vertical spring with spring constant k. The maximum force experienced by Grace's legs (when the trampoline is at its lowest point) is n times larger than Grace's weight. Ignore air resistance. 7 m h k 25.2kg 0.896m ✓ (no answer) Correct Answer: 2.47 158 N m n
Transcribed Image Text:Trampoline PE Grace has a mass of m. She drops a distance of h from a platform onto a trampoline which acts like a vertical spring with spring constant k. The maximum force experienced by Grace's legs (when the trampoline is at its lowest point) is n times larger than Grace's weight. Ignore air resistance. 7 m h k 25.2kg 0.896m ✓ (no answer) Correct Answer: 2.47 158 N m n
Solution: This solution uses a reference frame with y = 0 at the initial level of the trampoline. Conservation of Energy with (initial=before drop) and (final=trampoline at lowest point, stretched down a distance x): PE¿ + KE¿ + Wnc =PEf+KEf mg(h)+0+0 = mg(−x) + ¾½k(x)² +0 (Eqn. 1) The variable x is not in the problem statement. It can be related to the variable n by applying Hooke's Law at the trampoline's lowest point: |Flegs| = kx = n(mg) x = (mg) n Plug into (Eqn. 1) and rearrange terms: (mg)2 n² - 2k (mg)2 k :) n n-mgh = 0 2k Clean up by multiplying by (mg)2 : n² – 2n − ( 2kh = 0 mg
Transcribed Image Text:Solution: This solution uses a reference frame with y = 0 at the initial level of the trampoline. Conservation of Energy with (initial=before drop) and (final=trampoline at lowest point, stretched down a distance x): PE¿ + KE¿ + Wnc =PEf+KEf mg(h)+0+0 = mg(−x) + ¾½k(x)² +0 (Eqn. 1) The variable x is not in the problem statement. It can be related to the variable n by applying Hooke's Law at the trampoline's lowest point: |Flegs| = kx = n(mg) x = (mg) n Plug into (Eqn. 1) and rearrange terms: (mg)2 n² - 2k (mg)2 k :) n n-mgh = 0 2k Clean up by multiplying by (mg)2 : n² – 2n − ( 2kh = 0 mg
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