Block A weight `20 kg` placed on a smooth surface. Weight B of `2 kg` is mounted on the block . The coefficient of friction between the block and the weight is `0.25` Calculate the acceleration of the block and the weightand also the frictional force between the block and the weight when a horizontal force `2 N` applied to the weight as shown in fig What will these quantities be if the horizontal force is `20 N? (g = 10ms^(-2))`

Answer

`(1)/(11)ms^(-2):(20)/(11) N:7.5 ms^(-2) 0.25 ms^(-2), 5N`

Explanation

So long as the applied force does not product the limiting frictional force , the two bodies will have an relative motion , than is the two will behave an relative motion , that is the will behave like a compact body.
Here limiting friction `= 0.25 xx 2 xx 10 = 5 N`
Let `F` be the force that produces limiting friction force between the two .Than `F- 5 = 2a and 5 = 20a` where a is common acceleration Hence `F = 5.5 N` here the applied force is only `2 N` Hence the two bodes will behave as one and will have a common acceleration given by
`2 = (20 + 2)a or a = (1)/(11)ms^(-2)`
Now `f_(fr) = 20 xx (1)/(11) = (20)/(11) N`
where `F = 20 N` the applied force axceed the force required to produce the limiting friction force so there will be relative motion between the two frictional force = limiting frictional force`= 5 N`
From the free body diagram of the weight and the block.
`20 - 5 = 2a_(1)` or `a_(1) = 7.5 ms^(-2)`
`5 = 20a_(2)` or `a_(2)= 0.25 ms^(-2)`