A mass on a horizontal table is attached by a thin stringthat passes over a fric

Question

A mass on a horizontal table is attached by a thin stringthat passes over a frictionless, massless pulley to a 2.5 kg mass that hangs over the sideof the table 1 m above the ground. The system is released fromrestand the 2.5 kg mass strikes the ground at 0.9 s later. Thesystem is now placed in its initial position and a 1.1 kg massis placed on top of the block of mass m1. Releasedfrom rest, the 2.5 kg mass now strikes the ground 1.7 slater. The acceleration of gravity is 9.81 m/s2 .
Find the mass m1 on the table. Answer in units ofkg.
Determine the coefficient of kinetic friction between m1and the table. A mass on a horizontal table is attached by a thin stringthat passes over a frictionless, massless pulley to a 2.5 kg mass that hangs over the sideof the table 1 m above the ground. The system is released fromrestand the 2.5 kg mass strikes the ground at 0.9 s later. Thesystem is now placed in its initial position and a 1.1 kg massis placed on top of the block of mass m1. Releasedfrom rest, the 2.5 kg mass now strikes the ground 1.7 slater. The acceleration of gravity is 9.81 m/s2 .
Find the mass m1 on the table. Answer in units ofkg.
Determine the coefficient of kinetic friction between m1and the table.

Explanation / Answer

The solution to a similiar problem can be found here Let coefficient of friction between table and block be .    So, force of friction = f = m1g     In second case, f' = (m1+1.1)g     Now,    writing the force equations,     T - f = m1a     2*g - T = 2*a     T' - f' = (m1+1.1)*a     2*g - T' = 2*a'     Also, s = 1/2*a*t*t = 1/2*a'*t'*t' = 2.5m    From the above two you get a and a'.    Solve all these eqns to get theanswer.    2*9.8 - *m1*9.8 = (2+m1)*a => = 2 - (2+m1)*a/[m1*9.8] And,   2*9.8 - *(m1+1.1)*9.8 = (1.1+2+m1)*a' => = 2 - (3.1 +m1)*a'/[(m1+1.1)*9.8]    equating from the two equations,    (3.1+m1)*a'/(m1+1.1) = (2+m1)*a/(m1)    So, m1 = ?    So, force of friction = f = m1g     In second case, f' = (m1+1.1)g     Now,    writing the force equations,     T - f = m1a     2*g - T = 2*a     T' - f' = (m1+1.1)*a     2*g - T' = 2*a'     Also, s = 1/2*a*t*t = 1/2*a'*t'*t' = 2.5m    From the above two you get a and a'.    Solve all these eqns to get theanswer.    2*9.8 - *m1*9.8 = (2+m1)*a => = 2 - (2+m1)*a/[m1*9.8] And,   2*9.8 - *(m1+1.1)*9.8 = (1.1+2+m1)*a' => = 2 - (3.1 +m1)*a'/[(m1+1.1)*9.8]    equating from the two equations,    (3.1+m1)*a'/(m1+1.1) = (2+m1)*a/(m1)    So, m1 = ? Solve th equation for m1 Source: /physics-topic-5-484828-cpi0.aspx

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