Asked by Hanu
(a) A 10 kg box of candy rests on a floor with a coefficient of static friction of 0.30. What force is needed to move the box?
b) What is the coefficient of kinetic friction between a rubber tire and the road if a 2000 kg car needs 1.57 × 104 N to keep the car moving at a constant speed?
b) What is the coefficient of kinetic friction between a rubber tire and the road if a 2000 kg car needs 1.57 × 104 N to keep the car moving at a constant speed?
Answers
Answered by
Damon
normal force = m g = 10*9.81
so
friction force max = .3 * 10 * 9.81
which is min horizontal force required to start acceleration
F = m a
a = 0
so
friction Force - pulling force = 0
mu * 2000 * 9.81 = 1.57 * 10^4
so
friction force max = .3 * 10 * 9.81
which is min horizontal force required to start acceleration
F = m a
a = 0
so
friction Force - pulling force = 0
mu * 2000 * 9.81 = 1.57 * 10^4
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