Interpret the figure/data

In the standard version of this question, the three blocks on a smooth horizontal surface have masses: $A=2\,\text{kg},\quad B=3\,\text{kg},\quad C=5\,\text{kg}$ and they are pulled by an external force $F=80\,\text{N}.$

Let:

• $T_1$ = tension between blocks $A$ and $B$
• $T_2$ = tension between blocks $B$ and $C$

Find the acceleration of the system

Since the surface is smooth, the whole system moves together with the same acceleration.

Total mass: $M=2+3+5=10\,\text{kg}$

Therefore, $a=\frac{F}{M}=\frac{80}{10}=8\,\text{m/s}^2.$

Find $T_1$

Consider block $A$ alone. The only horizontal force on $A$ is the tension $T_1$.

So, $T_1 = m_A a = 2\times 8 = 16\,\text{N}.$

Find $T_2$

Consider the subsystem consisting of blocks $A+B$ together. The force pulling this subsystem is $T_2$.

Hence, $T_2=(m_A+m_B)a=(2+3)\times 8=40\,\text{N}.$

Compare with the given options

We get: $T_1=16\,\text{N},\quad T_2=40\,\text{N}.$

This does not match any of the listed options.

Check against stored answer

Stored correct answer is A: $40\,\text{N}, 64\,\text{N}$.

That pair does not follow from the usual interpretation unless the figure/masses are different from what is visible in the text. Since the actual figure is missing, the numerical result depends critically on the masses and the point of application of the 80 N force.

If one assumes the figure indeed corresponds to a different mass arrangement, then option A may be correct, but from the text alone and the standard mass set, I cannot derive it.

Therefore, based on the available information, I disagree with the stored answer because the figure is necessary to compute the tensions uniquely.