91Ó°ÊÓ

The masses of the two blocks, M = 2.0 kg and 2M

The spring constant of the spring, k = 200 N/m

The height of fall, h = 0.090 m

We can find the kinetic energy of the system and the hanging mass using the law of conservation energy. The maximum distance traveled by the hanging mass can be calculated by putting the velocity of it equal to zero in the equation obtained in part a for energy conservation.

Formulae:

The potential energy at a height, PE = mgh (1)

The kinetic energy of the body, $\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$ (2)

From energy conservation of the system, ${\mathrm{U}}_{\mathrm{i}}+\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}={\mathrm{U}}_{\mathrm{f}}+\mathrm{K}.\mathrm{E}{.}_{\mathrm{f}}$ (3)

Let the initial P.E. of the hanging block is zero and the upward direction is positive.

Using equation (1), the combined kinetic energy of the blocks using equations (1) and (2) is given as:

$$\begin{gathered} 0\mathrm{J}+0\mathrm{J}=\mathrm{mgh}+\frac{1}{2}{\mathrm{kx}}^2+\mathrm{K}.{\mathrm{E}}_{\mathrm{f}} \\ 0\mathrm{J}=2\left(2.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(-0.090\mathrm{m}\right)+\frac{1}{2}\left(200\mathrm{N}/\mathrm{m}\right){\left(0.090\mathrm{m}\right)}^2+\mathrm{K}.\mathrm{E}{.}_{\mathrm{f}} \\ \mathrm{K}.\mathrm{E}{.}_{\mathrm{f}}=2.7\mathrm{J} \end{gathered}$$

Therefore, the combined kinetic energy of two blocks is $2.7\mathrm{J}$.

From part a) we can write that, the kinetic energy of a mass is:

$$\begin{gathered} \frac{1}{2}\left(3\mathrm{M}\right){\mathrm{v}}^2=2.7\mathrm{J} \\ \frac{1}{2}{\mathrm{Mv}}^2=0.9\mathrm{J} \end{gathered}$$

The kinetic energy of the hanging mass is, given using the above equation:

$$\begin{gathered} \frac{1}{2}\left(2\mathrm{M}\right){\mathrm{v}}^2=2×0.9\mathrm{J} \\ =1.8\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy of the hanging block is 1.8 J

When the hanging block stops, the law of conservation of energy gives the maximum distance of the fall which is given using equations (1), (2), and (3) as follows:

$$\begin{gathered} 0\mathrm{J}+0\mathrm{J}=\left(2\mathrm{M}\right)×\mathrm{g}×\left(-\mathrm{d}\right)+\frac{1}{2}×\mathrm{k}×{\left(\mathrm{d}\right)}^2+0\mathrm{J} \\ \left(2\mathrm{M}\right)×\mathrm{g}×\left(-\mathrm{d}\right)+\frac{1}{2}×\mathrm{k}×{\left(\mathrm{d}\right)}^2=0 \\ 2\left(2.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(-\mathrm{d}\right)+\frac{1}{2}\left(200\mathrm{N}/\mathrm{m}\right){\left(-\mathrm{d}\right)}^2=0 \\ \mathrm{d}=0.39\mathrm{m} \end{gathered}$$

Therefore, the maximum distance through which the hanging block falls before momentarily stopping is 0.39 m