Question

Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1".The blocks sit on a frictionless surface. How do the magnitudes of the tensions in the rope compare?a. T1 > T2 b. T1 < T2 c. T1 = T2

Answer 1

Answer:

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$.

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 $a_{M_1}= a_{M_2} = a$

From the second Law of Newthon we have that the force applied for the mass $M_2$ is $F_{M_2}= M_2 a$ and we know that the force acting on the x axis for the mass 2 is $F_{M_2}= T_2$ so then we have that $T_2= M_2 a$

Now when we consider the system of $M_1 +M_2$ as a whole mass, this system have the same acceleration $a$ and on this case we will see that the only force acting on the entire system would be $T_1$ and then by the second law of Newton we have that:

$F_{M_1 +M_2} = T_1 = (M_1 +M_2) a$

And then if we compare $T_1$ and $T_2$ we see that :

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$.

So the best option for this case would be:

a. T1 > T2