An ideal spring hangs from the ceiling. a 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m from its original length when it reaches equilibrium. the mass is then lifted up a distance l = 0.0235 m from the equilibrium position and released. what is the kinetic energy of the mass at the instant it passes back through the equilibrium position?

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

Where, f = restoring force

Put the value into the formula

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

Here,

Here,

Put the value into the formula

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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