Question

Question #b3546

Answer 1

We need Hooke's force law `F=kx`, where `F` is the force applied, `k` is the spring constant, and `x` is the extension of the spring (in this case with 3kg it is 12cm) and the force of gravity `F=mg`,

Explanation:

Setting `m=3`kg and `m'=4`kg and `x=12`cm, we first have that
`mg=kx`. Recall that the spring potential energy is `E_s=1/2kx^2`, while the variation in gravitational potential energy is `E_g=-mgx` (minus because the direction of motion is downward here).

To find the stopping point with the new mass, the easiest approach is to consider the energies involved. The mass initially has zero kinetic energy and some gravitational potential energy, it then starts to fall turning the gravitational potential energy partly into kinetic energy and partly into spring potential energy. As the spring stretches it starts to absorb also the kinetic energy until the mass finally stops moving, when it thus has zero kinetic energy and some spring potential energy, having lost some gravitational potential energy.

Conservation of energy then tells us that `E_s+E_g+E_k` (where `E_k` is the kinetic energy) is constant. At the start and end pionts the mass is stationary and so `E_k=0`. So, we just have to consider the sum of the potential energies at the start and end.
We take `x=0` at the start and `E_g=0` there; we thus have

`0 + 0 = 1/2kx'^2 - m'gx'`, where `x'` is the end extension.

From `mg=kx`, we have `k=(mg)/x` and putting this all together we obtain

`x' = 2(m')/m x= 32`cm.