Question

Three blocks of masses ${m_1},{m_2}$ and ${m_3}$ are placed on a horizontal frictionless surface A force of $40N$ pulls the system then calculate the value of $T$ if ${m_1} = 10kg$ ,${m_2} = 6kg$ ,${m_3} = 4kg$

A) $40N$
B) $20N$
C) $10N$
D) $5N$

Answer 1

Hint:We can solve these types of questions by using Newton’s second law. First we find the net acceleration of whole system and after that we focus only on ${m_1}$ make free body diagram of this and again apply second law of Newton

Step by step solution:
As figure given in the question a force $F = 40N$ applied on ${m_1}$ due to this all blocks moving with a common acceleration $a$
First we calculate the common acceleration of the system by which these blocks are moving
We consider all three blocks as a system as shown in figure

Mass of system $M = {m_1} + {m_2} + {m_3} = 10 + 6 + 4 = 20kg$
External force on whole system $40N$
So apply Newton’s second law which states ${F_{ext}} = Ma$
Where $M \Rightarrow $ mass of the system
$a \Rightarrow $ Acceleration of system
$ \Rightarrow 40 = 20 \times a$
So common acceleration or net acceleration
$ \Rightarrow a = 2m/{s^2}$
Hence all the blocks having same acceleration $2m/{s^2}$
Now focus only on ${m_1}$
Diagram of ${m_1}$ given below

Force on ${m_1}$
An applied force $F = 40N$ and tension $T$
Apply Newton’s law ${f_{net}} = ma$
$ \Rightarrow F - T = {m_1} \times a$
$ \Rightarrow 40 - T = 10 \times 2$
Solving this
$ \Rightarrow T = 40 - 20$
$\therefore T = 20N$
Hence tension $T = 20N$

Option B is correct

Note:We used here Newton’s second law which states that the rate of change of momentum of a body is directly proportional to applied force
$
   \Rightarrow F = \dfrac{{dP}}{{dt}} \\
   \Rightarrow F = \dfrac{{dmv}}{{dt}} \\
   \Rightarrow F = m\dfrac{{dv}}{{dt}} \\
 $
We know $\dfrac{{dv}}{{dt}} = a$
$ \Rightarrow F = ma$
If a force $F$ acting in a body of mass $m$ then it starts moving with an acceleration $a$ can written as $F = ma$