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)` )

Options

A. 8N
B. 10 N
C. 6N
D. 4N

Answer

A

Explanation

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

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

`f =mu R_(1) = 0.5 xx 2g = 0.5 xx 2xx 10 `
`f = 10 N`
Since the static friction on 2 kg block is more than force applied on it.
i.e., `f gt 2N`
Hence 2kg body will move along the direction of 4kg body.
Hence net friction force on the block of 2kg
`f= f-2N `
`=10 -2 =8 N`