In the Fig 7.89 shown , `m_(2) = 2.5` kg , h = `1.5` m , the system is released from rest at t = 0 and the mass `m_(2)` strikes ground at t = `0.82`s . The system is placed in its initial position and a `1.2` kg mass is placed on top of the block of mass `m_(1)` . Released from rest , the mass `m_(2)` now strikes the ground `1.3` s later . Determine the mass `m_(1)` and the coefficient of kinetic friction between `m_(1)` and the shelf .

Answer

`1.22 ` kg ; `0.672`

Explanation

Initial acceleration of system when `1.2` kg mass is not placed over `m_(1)` kg block

`a_(0) = sqrt((2h)/(t^(2))) = sqrt((2 xx 1.5)/(0.82 xx 0.82))`
As , `" " m_(2) g - T = m_(2)a_(0) " " …… (i) `
`T - mu m_(1) g = m_(1) a_(0) " " ...... (ii)`
Adding Eqns. (i) and (ii)
`g (m_(2) - mum_(1)) = a_(0) (m_(1) + m_(2)) " " ..... (iii)`
Now when `1.2` kg mass is placed , then
`a= sqrt((2h)/(t'^(2))) = sqrt((2 xx 1.5)/(1.3 xx 1.3))`
as `m_(2)g - T = m_(2)a " " ...... (iv)`
`T - mu(m_(1) + 1.2)g = a [m_(2) + m_(1) + 1.2] " " .....(v)`
Adding Eqns. (iv) and (v)
`m_(2)g - mu_(g) (m_(1) + 1.2) = a[m_(2) + m_(1) + 1.2] " " .... (vi)`
Solving Eqns. (iii) and (vi) ,
we get values of `mu` and `m_(1)`.