Answer:
[tex] (M_1 + M_2) a > M_2 a[/tex]
Becuase [tex] M_1 +M_2> M_2[/tex]
So then we can conclude that:
[tex] T_1 > T_2[/tex]
And that makes sense since the force [tex] T_1[/tex] needs to accelerate the two masses and [tex] T_2[/tex] just need to accelerate [tex] M_2[/tex].
So the best option for this case would be:
a. T1 > T2
See explanation below.
Explanation:
For this case we consider the system as shown on the figure attached.
Since the system is connected the acceleration for both masses are equal, that is [tex] a_{M_1}= a_{M_2} = a[/tex]
From the second Law of Newthon we have that the force applied for the mass [tex] M_2[/tex] is [tex] F_{M_2}= M_2 a[/tex] and we know that the force acting on the x axis for the mass 2 is [tex] F_{M_2}= T_2[/tex] so then we have that [tex] T_2= M_2 a[/tex]
Now when we consider the system of [tex] M_1 +M_2[/tex] as a whole mass, this system have the same acceleration [tex]a[/tex] and on this case we will see that the only force acting on the entire system would be [tex] T_1[/tex] and then by the second law of Newton we have that:
[tex] F_{M_1 +M_2} = T_1 = (M_1 +M_2) a[/tex]
And then if we compare [tex] T_1[/tex] and [tex] T_2[/tex] we see that :
[tex] (M_1 + M_2) a > M_2 a[/tex]
Becuase [tex] M_1 +M_2> M_2[/tex]
So then we can conclude that:
[tex] T_1 > T_2[/tex]
And that makes sense since the force [tex] T_1[/tex] needs to accelerate the two masses and [tex] T_2[/tex] just need to accelerate [tex] M_2[/tex].
So the best option for this case would be:
a. T1 > T2
