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
$$\begin{gathered} \Rightarrow F={\displaystyle \frac{dP}{dt}} \\ \Rightarrow F={\displaystyle \frac{dmv}{dt}} \\ \Rightarrow F=m{\displaystyle \frac{dv}{dt}} \end{gathered}$$
We know ${\displaystyle \frac{dv}{dt}}=a$
$\Rightarrow F=ma$
If a force $F$ acting in a body of mass $m$ then it starts moving with an acceleration $a$ can written as $F=ma$