Question

Pulling Two Blocks In the situation shown in the figure, a person is pulling with a constant, nonzero force F⃗ on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass. (Figure 1)

Part A Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B? Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B? SubmitHintsMy AnswersGive UpReview Part Part B How does the magnitude of the tension in string 1, T1 , compare with the tension in string 2, T2 ? How does the magnitude of the tension in string 1, \texttip{T_{\rm 1}}{T_1}, compare with the tension in string 2, \texttip{T_{\rm 2}}{T_2}? Block A has a larger acceleration than block B. Block B has a larger acceleration than block A. Both blocks have the same acceleration. More information is needed to determine the relationship between the accelerations. SubmitHintsMy AnswersGive UpReview Part Provide FeedbackContinue T1>T2 T1=T2 T1<T2 More information is needed to determine the relationship between T1 and T2 . Figure 1 of 1

Answer 1

Concept and reason

The concept of acceleration due to applied force is used to find the acceleration of each box.

The formula for the force applied on the body and tension on the string is used to find the tension on each string.

Fundamentals

According to the relation between force and acceleration the acceleration of the object depends upon the mass of the object and force applied.

The formula for the force acting on a body and its acceleration is given as:

$F = ma$

Here, m is the mass and a is the acceleration of the body moving forward.

In case if two objects are tied by a string and pulled in single direction then force acts on both objects.

(A)

The formula for the force acting on a body and its acceleration is given as:

$F = ma$ .

Or rearranging this equation we have;

$a = \frac{F}{m}$

In this case we have two blocks of different sizes.

Let us assume both have different masses such as block A has mass of ${m_1}$ and block B has a mass of ${m_2}$ .

In this case when force is acting on both blocks in same direction the force is acting on the blocks collectively. Hence we can consider these masses as single mass that is; ${m_1} + {m_2}$ .

So the acceleration of the system of the blocks would be given by the formula $a = \frac{F}{m}$ ;

As, $a = \frac{F}{{{m_1} + {m_2}}}{\rm{m}}{{\rm{s}}^{ - 2}}$ .

Hence we can see the acceleration of the whole system is same. Thus the acceleration of both blocks is same.

Part A

[Part A]

(B)

Let the tension in string 1 is given by ${T_1}$ and tension on the string 2 is given by ${T_2}$ .

From the image it can be clearly observed that string 1 is attached to two masses ${m_1}$ and ${m_2}$ .

And the string 2 is attached to only mass ${m_2}$ .

By comparing the mass attached to both strings we can clearly compare that the tension on the string that is connected to less mass would be lesser as compared to other string.

As it is clear that ${m_2} < {m_1} + {m_2}$ , hence the tension on the string 2 that is ${T_2}$ will be less than tension on the string 1 that is ${T_1}$ .

The relation between tensions on both strings is ${T_1} > {T_2}$ .

Ans: Part A

The acceleration of both blocks is same.