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. . .

e) Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its lowest point

f) Calculate the kinetic energy of the cat when the cat is at its lowest point.        

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

h) Calculate the sum of these three energies when the cat is at its lowest point

Explanation / Answer

We now that the elasatic potential energy is given by

Energy =(1/2)kA2

But here at the lowest point, the amplitude of the spring becomes twice the amplitude A =2A

Now Energy =(1/2) *k*(2A)2 =2kA2

f and g)

The kinetic energy of the cat when the cat is at its lowest point is zero, because the velocity of the cat the lowest point is zero . Simlalry at the lowest point height the cat's potential energy is also zero.

h)

The sum of these three energies when the cat is at its lowest point is given by the conservation of energy energy at lowest point is equal to energy at the heighest point.

Total energy energy when the cat at the lowest point is

(TE) =KE+PE+ Elastic PE

=0+0+mg(2A)

=2mgA

=2*4.00*9.81*0.050

=3.924J

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