3. The coefficient of kinetic friction between the surface and the blocks is 0.12. If F = 110N and m =49 kg in the figure, how much does friction act on m? What is the acceleration of m? g = 9.81 m/s" .12 = Friction F=110N Friction=.12 fg Friction m= 49 Kg Fnet=110+.12 Acceleration: acal= 110.12 / 49 = 2.2447 = 2.25
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12 * F_n Friction = 0.12 * 49 kg * 9.81 m/s^2 Friction = 58.03 N Show more…
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