Given `k_1=1500Nm^-1`, `k_2=500Nm^-1`, `m_1=2kg`, `m_2=1kg`. Find:

a. potential energy stored in the spring in equilibrium, and
b. work done in slowly pulling down `m_2` by `8cm`.

Answer

a. `0.4J`; b `1J`;

Explanation

Let the initial extension in the springs of force constants `k_1` and `k_2`, at equilibrium position, be `x_1` and `x_2`. Then
`x_2=(m_2g)/(k_2)`, `x_1=((m_1+m_2)g)/(k_1)`

a. Potential energy stored in the springs in equilibrium position is
`U_1=1/2k_1x_1^2+1/2k_2x_2^2`
Putting values of `x_1` and `x_2` from above, we get `U_1=0.4J`
b. Let `Deltax_1` and `Deltax_2` be additional elongations caused by pulling `m_2` by `l=8cm`. Additional forces on `m_1` are equal and in opposite directions.
`impliesk_1Deltax_1=k_2Deltax_2` (i)
Also, `Deltax_1+Deltax_2=l` (ii)
`Deltax_1` and `Deltax_2` can be found from Eqs. (i) and (ii).
From work-energy theorem,
`w_g+w_p+w_s=0`
(where `w_p` is the work done by the pulling force)
`impliesw_p=-w_s-w_g=(U_2-U_1)-[m_1gDeltax_1+m_2g(Deltax_1+Deltax_2)]`
where `U_2=1/2k_1(x_1+Deltax_1)^2+1/2k_2(x_2+Deltax_2)^2`
Putting the values, we get
`impliesw_p=1J`