We are given three blocks, each with a mass of 434.3 g (which is 0.4343 kg), being pulled up by a force of 7 N. We need to determine the tension between the top and middle block.

Step-by-Step Solution:

1. Mass of each block:

   • Convert the mass from grams to kilograms: $m = 434.3 \, \text{g} = 0.4343 \, \text{kg}$

2. Total mass of the three blocks:

   • Since all three blocks have the same mass, the total mass is: $M_{\text{total}} = 3 \times 0.4343 \, \text{kg} = 1.3029 \, \text{kg}$

3. Acceleration of the system:

   • The net force $F_{\text{net}}$ applied is 7 N, and the total mass being pulled is 1.3029 kg.
   • To find the acceleration of the system, use Newton’s second law: $F_{\text{net}} = M_{\text{total}} \times a$ Solving for acceleration $a$: $a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{7}{1.3029} \approx 5.37 \, \text{m/s}^2$

4. Forces on the top two blocks:

   • The top two blocks consist of the top block (mass $m$) and the middle block (mass $m$), which together have a combined mass of $2m = 2 \times 0.4343 = 0.8686 \, \text{kg}$.
   • To find the tension $T$ between the top and middle blocks, we need to calculate the net force required to accelerate the bottom two blocks upwards. This net force is what the top block is pulling with, i.e., the tension.

   Using Newton’s second law again: $T = 2m \times a$ Substituting the values: $T = 0.8686 \times 5.37 \approx 4.66 \, \text{N}$

Thus, the tension between the top and middle block is approximately 4.66 N.

Would you like any further details or clarification?

---

Here are 5 related questions to explore:

1. How would the tension change if the force was increased to 10 N?
2. What if the masses of the blocks were different?
3. How would friction affect the tension in the string?
4. What is the significance of using a massless string in this problem?
5. How does the acceleration affect the net force on each block?

Tip: Always convert units to SI (e.g., grams to kilograms, newtons, etc.) before calculations for consistency and accuracy.