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physics kinetic friction?
A 15.9 kg box rests on a horizontal surface. The coefficient of kinetic friction between the surface and the box is 0.33. The box is pulled to the right by a force of 109 N.
What is the acceleration in m/s^2?
- AshLv 72 months agoFavorite Answer
F(net) = F(applied) - F(friction)
ma = F(applied) - μmg
a = [F(applied) - μmg]/m
a = [(109 N) - (0.33)*(15.9 kg)(9.8 m/s²)]/(15.9 kg)
a = 3.6 m/s²
- oubaasLv 72 months ago
acceleration a = (F-FF)/m = (109-15.9*9.806*0.33)/15.9 = 3.6 m/sec^2 (2 sign. digits)
- WhomeLv 72 months ago
F = ma
109 - 0.33(15.9)(9.81) = 15.9a
a = 3.6180... ≈ 3.62 m/s² ◄
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