Question

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?

Answer 1

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

Step-by-step 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

`F= mg`

Where, f = restoring force

`kx=mg`

`k=(mg)/(x)`

Put the value into the formula

`k=(2.15*9.8)/(0.0895)`

`k=235.41\ N/m`

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

`K.E=(1)/(2)mv^2`

Here,
`v = A\omega`

`K.E=(1)/(2)m*(A\omega)^2`

Here,
`\omega=\sqrt{(k)/(m)}^2`

`K.E=(1)/(2)m* A^2\sqrt{(k)/(m)}^2`

`K.E=(1)/(2)kA^2`

Put the value into the formula

`K.E=(1)/(2)*235.41*(0.0235)^2`

`K.E=0.06500\ J`

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