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 $4 \,kg$ block is placed on a smooth horizontal surface.) (Acceleration due to gravity $=10 \,ms ^{-2}$.)

• 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 $2 \,kg$ and $4 \,kg$ block then the static friction force on $2 kg$ block.

$f'=\mu R_{1}=0.5 \times 2 g =0.5 \times 2 \times 10 [$ Given, $\mu=0.5]$
$f'=10 \,N$
Since, the static friction on $2 \,kg$ block is more than force applied on it.
i.e, $f'>2\, N$
Hence, $2 \,kg$ body will move along the direction of $4\,kg$ body.
Hence, net friction force on the block of $2 kg$,
$f=f'- 2 \,N$
$=10-2=8\, N$