Question

Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings as shown on a frictionless table. They are pulled with a force of $40\, N$. If $m_{1}=10\, kg , m_{2}=6 \,kg$ and $m_{3}=4\, kg$, then tension $T_{2}$ will be

• A. 10 N
• B. 20 N
• C. 32 N
• D. 40 N

Answer 1

Answer: (B)

From Newton's law, the force $(F)$ is given by
$F=M a$
where $a$ is acceleration and $M$ the mass.
Given, $ M =m_{1}+m_{2}+m_{3}$
$=10+6+4=20 \,kg , $
$ F =40\, N $
$ \therefore a =\frac{F}{M}=\frac{40}{20}=2 \,m / s ^{2} $
Tension between $10\, kg$ and $6 \,kg$ masses is given by
$T_{2} =\left(m_{2}+m_{3}\right) a$
$=(6+4) 2$
$=10 \times 2=20\, N$