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 a simple harmonic motion. The amplitude is 0.050 m, and at the highest point of the motion, the spring has its natural unstretched length.
What is the force constant of this spring?
What is the magnitude of the cat's velocity at the equilibrium position of the cat's motion?
What is the force constant of this spring?
What is the magnitude of the cat's velocity at the equilibrium position of the cat's motion?
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Hello
if the amplitude is 0.05 m, and the uncomressed length is equal to the upper point of the oscillation,
then the cat' compresses the spring by also by 0.05 m.
and k = F/s = 4kg*9.81m/s^2 N/ 0.05 m = 784.8 N/m <--- ans
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the total mechanical energy of the system is the spring energy at the top and bottom point of the oscillation: E = 1/2*k*s^2
E = 1/2*784.8*0.05^2
E = 0.981 Nm
This spring energy is fully converted into kinetic energy at the equilibrium position:
0.981 Nm = 1/2*mv^2
v^2 = 2*0.981/4
v = 0.7 m/s <--- ans.
Regards
if the amplitude is 0.05 m, and the uncomressed length is equal to the upper point of the oscillation,
then the cat' compresses the spring by also by 0.05 m.
and k = F/s = 4kg*9.81m/s^2 N/ 0.05 m = 784.8 N/m <--- ans
---------
the total mechanical energy of the system is the spring energy at the top and bottom point of the oscillation: E = 1/2*k*s^2
E = 1/2*784.8*0.05^2
E = 0.981 Nm
This spring energy is fully converted into kinetic energy at the equilibrium position:
0.981 Nm = 1/2*mv^2
v^2 = 2*0.981/4
v = 0.7 m/s <--- ans.
Regards