Question

Two blocks, of masses M = 2.0 kgand 2M, are connected to a spring of spring constantk = 200 N/mthat has one end fixed, as shown in Fig. 8-69. The horizontal surface and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen 0.090m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.090m ?(c) What maximum distance does the hanging block fall before momentarily stopping?

Short answer

1. The combined kinetic energy of two blocks is 2.7 J
2. The kinetic energy of the hanging block is 1.8 J
3. The maximum distance through which the hanging block falls before momentarily stopping is 0.39 m

Step-by-step solution

The given data

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

Understanding the concept of energy

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)

a) Calculation of the combined kinetic energy

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 $2.7\mathrm{J}$.

b) Calculation of the kinetic energy of the hanging block

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:

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

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

c) Calculation of the maximum distance of the fall

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:

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

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