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

Part A.) Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its highest point.

The answer is Uspring=0 J

Part B.) Calculate the kinetic energy of the cat when the cat is at its highest point.

The answer is K=0 J

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

The answer is Ugrav= 3.92 J

Part D.) Calculate the sum of these three energies when the cat is at its highest point.

The answer is E= 3.92 J

Part 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.

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

K= ?? J

PArt 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.

Ugrav=?? J

Part H.) Calculate the sum of these three energies when the cat is at its lowest point.

E=?? J

Explanation / Answer

mass = m =4.00 kg

amplitude = A = 0.050 m

(A) when cat is at highest point, spring is unstretched, so the elastic potential energy of the spring is zero.

(B) At highest point, velocity of cat is zero, so kinetic energy is zero too

(C) Potential energy = mgh = mg(2A) = 4.00 x 9.81 x 2 x 0.050 = 3.92 J

(D) The sum of these three energies when the cat is at its highest point =E = 0 + 0 + 3.92 = 3.92 J

(E) Elastic potential energy at lowest point= ?

by conservation of mechanical energy,

(KE + PE +EPE) at lowest point = (KE + PE +EPE) at highest point

0 + 0 + EPE = 0 + 3.92 + 0

EPE = 3.92 J

F) KE = 0 at lowest point as cat is instantaneously at rest at this position

G) At lowest point Gravitational potential energy is also zero

H) E = EPE + KE +  Gravitational potential energy = 3.92 + 0 + 0 = 3.92 J

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