A 4.0 kg block is put on top of a 5.0kg block. In order to cause the top block to slip on the bottom one, a horizontal force of 12N must be applied to the top block. Assuming a friction less table the maximum horizontal force F which can be applied to the lower block so that the blocks, will move together should be

Options

A. `lt30N`
B. `lt27N`
C. `lt15N`
D. `lt9N`

Answer

B

Explanation

As both the blocks move together, their accelration
(a) should be same.
For upper block of 4 kg.
force (f)=12N
`therefore4xxa=12`
or `a=(12)/(4)=3 m//s^(3)`
Maximum horizontal force on lower block of 5 kg is given by `F-f=(5)xx(3)`
or `F-12=5xx3` or `F=15+12v=27N`
`therefore` Maximum horizontal force =27N
`therefore` force should be less than 27 N