In the Fig 7.89 shown , m_(2) = 2.5 kg, h = 1.5 m , the system is released fromrest at t = 0 and the mass m_(2) strikes ground at t = 0.82s . The system is placed in its initial position and a 1.2 kgmass 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 .

In the Fig 7.89 shown , m_(2) = 2.5 kg, h = 1.5 m , the system is rele

1.	In the Fig 7.89 shown , m_(2) = 2.5 kg, h = 1.5 m , the system is released fromrest at t = 0 and the mass m_(2) strikes ground at t = 0.82s . The system is placed in its initial position and a 1.2 kgmass 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» Solution :Initial acceleration of system when `1.2` kg mass is not placedover `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)/(1.3xx 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)`.