A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is  (b) at its lowest point

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is  (b) at its lowest point

 

found the answer on quizlet but i cannot understand the Us part. it says Us= mgh yet there is a times 2 before 0.05 (check picture for context)

 

please explain it to me. thanks

(b) When the cat is at its lowest point, then at the lowest
point of the motion the spring has its maximum stretched length i.e. æ = 24, so from equation (1), the elastic
potential energy of the spring will be
Us = k (24)°
(8)
= 2k A?
But here we know
mg = kr
= 2kA
From the above, we get
mg
%3D
2A
Hence equation (8) becomes
mg
Us = 29 A
A
= mg A
(9)
= 4 kg x 9.8 ms 2 x 2 x 0.05 m
= 3.92 J
-
Transcribed Image Text:(b) When the cat is at its lowest point, then at the lowest point of the motion the spring has its maximum stretched length i.e. æ = 24, so from equation (1), the elastic potential energy of the spring will be Us = k (24)° (8) = 2k A? But here we know mg = kr = 2kA From the above, we get mg %3D 2A Hence equation (8) becomes mg Us = 29 A A = mg A (9) = 4 kg x 9.8 ms 2 x 2 x 0.05 m = 3.92 J -
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