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4-1.Newton's Laws of Motion
normal
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
Acceleration of system $=\frac{\mathrm{F}_{\mathrm{net}}}{\mathrm{M}_{\mathrm{total}}}$
$=\frac{14}{4+2+1}=2 \mathrm{m} / \mathrm{s}^{2}$
The contact force between $4\, kg$ $\& 2$ $kg$ block will
move $2\, kg \,\&\, 1\, kg$ block with the same
acceleration
so $\mathrm{F}_{\text {contact }}=(2+1) \mathrm{a}=3(2)=6 \mathrm{N}$
$=\frac{14}{4+2+1}=2 \mathrm{m} / \mathrm{s}^{2}$
The contact force between $4\, kg$ $\& 2$ $kg$ block will
move $2\, kg \,\&\, 1\, kg$ block with the same
acceleration
so $\mathrm{F}_{\text {contact }}=(2+1) \mathrm{a}=3(2)=6 \mathrm{N}$
Standard 11
Physics
