Question

Objects with masses m1 = 8.0 kg and m2 = 5.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, m2 falls 1.00 m in 1.70 s, determine the coefficient of kinetic friction between m1 and the table.

Answer 1

first find the acceleration
distance(d), times(t), initial velocity(u) is given. find a?
use this equation : d = ut+(1/2)at^2
1 = 0 + 0.5a(1.7*1.7)
1 = 1.445a
a = 0.692 m/s^2

consider m2 :

ΣF = m2a
m2g-T = m2a
T = m2g-m2a = m2(g-a)
T = 5(9.8-0.692)
T = 45.54 N
consider m1 :

Σ F = m1a
T - F(friction) = m1a
F = T-m1a
μmg = 45.54 - (8*0.692)
μ*8*9.8 = 40
μ = 0.51

Answer 2

first find the acceleration
distance(d), times(t), initial velocity(u) is given. find a?
use this equation : d = ut+(1/2)at^2
1 = 0 + 0.5a(1.7*1.7)
1 = 1.445a
a = 0.692 m/s^2

consider m2 :

ΣF = m2a
m2g-T = m2a
T = m2g-m2a = m2(g-a)
T = 5(9.8-0.692)
T = 45.54 N
consider m1 :

Σ F = m1a
T - F(friction) = m1a
F = T-m1a
μmg = 45.54 - (8*0.692)
μ*8*9.8 = 40
μ = 0.51