Question

In the arrangement shown in the figure, the coefficient of friction between two blocks is $0.5$. The force of friction between the two blocks is (Assume that the $4kg$ block is placed on a smooth horizontal surface.) (Acceleration due to gravity $=10m{s}^{-2}$.)

• A. 8 N
• B. 10 N
• C. 6 N
• D. 4 N

Answer 1

Answer: (A)

According to the question, the given condition is shown in the figure,

If $f$ be the friction force between $2kg$ and $4kg$ block then the static friction force on $2kg$ block.

$f^{\prime }=\mu R_1=0.5\times 2g=0.5\times 2\times 10[$ Given, $\mu =0.5]$
$f^{\prime }=10N$
Since, the static friction on $2kg$ block is more than force applied on it.
i.e, $f^{\prime }>2N$
Hence, $2kg$ body will move along the direction of $4kg$ body.
Hence, net friction force on the block of $2kg$,
$f=f^{\prime }-2N$
$=10-2=8N$