Question

please answer......​

Answer 1

Let us assume that the maximum compression in the spring be x0x0. Then using the law of conservation of energy:

12mv/2+12kx/2=12kx/20...∗12mv/2+12kx/2=12kx/02...∗

Also, it is given that:

KineticKinetic EnergyEnergy == 2×Potential2×Potential energy of spring

12mv/2=2×12kx/2=kx/2 12mv/2=2×12kx2=kx/2

Therefore, from ,

kx/2+12kx/2=12kx/20kx × 2+12kx/2=12kx/02

⟹⟹ 32kx/2=12kx × 2032kx/2=12kx/02

⟹⟹ 3x/2=x203/x2=x/02

⟹⟹ x=x/203−−−√x=x/023

∴ x=x/03–√x=x/03

Thus, when the kinetic energy of spring is twice the elastic potential energy the compression of the spring will be ≈≈ 58%58%of the maximum compression

Answer 2

Answer:

AnsweR:

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