Determine the mass & coefficient of kinetic friction ?

Question

A mass m1 on a horizontal table is attached by a thin string that passes over a frictionless & massless pulley to 2.5 Kg mass m2 that hangs over the side of the table 1.5 Meter above the ground.The system is released from rest at t=0 & the 2.5 Kg mass m2 strikes the ground at t=0.82 Sec.
The system is now placed in its initial position & a 1.2 Kg mass is placed on top of the block of mass m1.
Released from rest , the 2.5 Kg mass now strikes the ground
1.3 Sec. later, Determine the mass m1 & the coefficient of kinetic friction m1 & the table.

Answer 1

You have to set up 2 equations to solve 2 unknowns: u (coeff of friction) and m1

For the first system, pulling force = m2g
Frictional force = um1g
Net force = (m2-um1)g
Net accel of the first system a1 = (m2-um1)g/(m1+m2)

The downward accel a of m2 is also determined by the time of flight

d = vot + 1/2 * a1* t^2, With vo = 0, you get
a1 = 2*1.,5/(0.82^2) = 4.46

Equating the two equations for a1,
4.46 = (2.5 - u*m1)g/(2.5+m1)

For the second system, the line of argument is the same.
Frictional force = u (m1 + 1.2)
Pulling force = m2 g (same as before)

a2 = (m2 - u(m1+ 1.2))g/(m1+1.2 + m2)

The flight time of 1.3 sec helps find a2,
a2 = 2* 1.5/(1.3^2) = 1.78

So 1.78 = (2.5 - u(m1+1.2))g/(m1 + 1.2 + 2.5)

You can solve for u and m1