When a paperweight and tray are placed together and put into a horizontal oscillating motion, the coefficient of static friction between them is 0.2
Given frequency of motion (f) = 1.22 Hz.
the amplitude of the motion (A) = 5.00 x 10^-2 m
The coefficients of the normal force applied by the surface are determined as friction force.
Force of friction (Fr) = μ x normal force (F)
coefficient(μ) = F/Fr
we know that Fr = kx = mω^
2A anf F = mg
mω^
2A = μmg
μ = ω^
2A/g and ω = 2πf
μ = (2x3.14x1.22)^2 x 5x10^-2/9.8 = 0.2
Hence the coefficient = 0.2
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