In the Fig. 7.85 shown a constant force is applied on the lower block , just large sliding out from between the upper block and the table . Determine the acceleration of each block .

Answer

`0.98 m//s^(2) , 1.96 m//s^(2)`

Explanation

For upper block

`f_(1) = 5 xx a_(1)`
`a_(1) = 0.98 m// sec^(2)`

For lower block

`F - mu_(k)N_(1) - mu_(k) N_(2) = M xx a_(2)`
`F - 0.1 xx 5 xx 9.8 - 0.4 xx 20 xx 9.8 = 15 xx a_(2) " " ..... (i) `
[Value of F = `(3)/(10) xx 5 xx 4.9 + (5)/(10) xx 20 xx 4.9`]
By putting this value of F in equation (i) , we get `a_(2)`