The figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a) μs = 0.600 and μk = 0.530 and (b) μs = 0.400 and μk = 0.330?
F cos(theta) = 0.500 cos(20) = 0.47 mg
A)
Static force, f = us N = us (mg - F sin(theta))
F = 0.6(mg - 0.500 mg sin(20)) = 0.497 mg
Since, F cos(theta) < f, box will remain stationary, acceleration will be zero
B)
Static friction force, f = 0.4 (mg - 0.5 sin(20)) = 0.33 mg
Acceleration, a = F cos(theta) - uk (mg - F sin(theta))/m
a = (0.5 x 9.8 cos(20)) - (0.33 x 9.8) - (0.5 x 9.8 x sin(20)) = 1.92 m/s^2
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