A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spri

Question

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

i) Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its equilibrium position.        

j) Calculate the kinetic energy of the cat when the cat is at its equilibrium position

k) Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at its equilibrium position.

L) Calculate the sum of these three energies when the cat is at its equilibrium position.  

Explanation / Answer

Solution:

Amplitude = A = 0.05 m

mass = m = 4 kg

i) When the cat hangs from the spring , the displacement at the lowest point is Twice the amplitude.

Weight of the cat = mg is the restoring force F

F= -kX

=> -mg = kA

=> spring constant k = mg/A

So Elastic potential energy = 1/2 kx^2 = 1/2 * (mg/A) ( 2A)^2 = 1/2 * mg* 4A^2/A = 2mgA

   => Elastic potential energy = U = 2mgA = 2* 4*9.8*(0.05) = 3.92 J

j) the kinetic energy at the equilibrium position = 0

k) Gravitational potential energy of the cat = mg = (4)(9.8) =39.2 J

l) The sum of these three energies = 3.92+0+39.2 = 43.12 J

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