Two blocks of masses M = 3 kg and m = 2 kg are in contact on a horizontal table. A constant horizontal force F = 5 N is applied to block M as shown. There is a constant frictional force of 2 N between the table and the block m but no frictional force between the table and the first block M, then acceleration of the two blocks is

Options

A. `0.4 ms^(-2)`
B. `0.6 ms^(-2)`
C. `0.8 ms^(-2)`
D. `1 ms^(-2)`

Answer

B

Explanation

If both blocks move together
`F - f = (m+M) a`
`therefore " "a=(F-f)/(m+M)`
`=(5-2)/(5)=(3)/(5)=0.6 m//s^(2)`