1) m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
2)m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
3) m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
4) m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m
Answer:
M2>M4>M3>M1
Explanation:
[tex]E=0.5(kx^{2}+mv^{2})[/tex]
For m1
[tex]E1=0.5*500*0.02^{2}=0.1 J[/tex] since v=0
For m2
[tex]E2=0.5*0.6*1^{2}=0.35 J[/tex] since x=0
For m3
[tex]E=0.5[(400*(0.01)^{2})+(1.2*0.5^{2})]=0.17 J[/tex]
For m4
[tex]E4=0.5[(200*0.05^{2})+(2*0.2^{2})]=0.29 J[/tex]
Therefore, M2>M4>M3>M1
The arrangement of total mechanical energy of the system from the largest to smallest depends on the mass, speed and displacement of each mass.
The given parameters;
The displacement of each block is calculated by applying the principle of conservation of mechanical energy as follows;
[tex]\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\x^2 = \frac{mv^2}{k} \\\\x = \sqrt{\frac{mv^2}{k} }[/tex]
The total mechanical energy at any point during the oscillation is given as;
[tex]E = K.E + U_x\\\\E = \frac{1}{2} mv^2 + \frac{1}{2} kx^2[/tex]
The total mechanical energy of each system depends on the mass, speed and displacement of each mass.
Thus, we can conclude that the arrangement of total mechanical energy of the system from the largest to smallest depends on the mass, speed and displacement of each mass.
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