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4-1.Newton's Laws of Motion
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Three blocks, $A, B$ and $C,$ of masses $4\,kg, 2\,kg$ and $1\,kg$ respectively, are in contact on a frictionless surface, as shown. If a force of $14\,\,N$ is applied on the $4\,\,kg$ block, then the contact force between $A$ and $B$ is ....... $N$
A$6$
B$8$
C$18$
D$2$
Solution
Here, $F=14 N$
$m_{A}=4 k g, m_{B}=2 k g, m_{C}=1 k g$
Total mass $m=m_{A}+m_{B}+m_{C}$
$=4+2+1=7 k g$
Acceleration of the system
$a=\frac{F}{m}=\frac{14}{7}=2 m / s^{2}$
The contact force between $4 k g$ and $1 k g$ block with the same acceleration Let
$F_{1}$ be the contactv force
$\therefore F_{1}=\left(m_{B}+m_{C}\right) a=(2+1)=6 N$
$m_{A}=4 k g, m_{B}=2 k g, m_{C}=1 k g$
Total mass $m=m_{A}+m_{B}+m_{C}$
$=4+2+1=7 k g$
Acceleration of the system
$a=\frac{F}{m}=\frac{14}{7}=2 m / s^{2}$
The contact force between $4 k g$ and $1 k g$ block with the same acceleration Let
$F_{1}$ be the contactv force
$\therefore F_{1}=\left(m_{B}+m_{C}\right) a=(2+1)=6 N$
Standard 11
Physics
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