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Applied force of 455N is applied to a 27.0kg bicycle, which causes the bicycle to accelerate at 3.60m/s2. What is the coefficient of kinetic friction between the brick tires and road?

 

6 years ago

Answered By Alexandria M

So looking at the free body diagram below we can write the summation of forces in the horizontal and vertical directions: Fx = Fapplied-Fkfriction = ma and Fy = Fn - Fg = 0. We know mass, acceleration, and Fapplied. We'll let muK be the coefficent of kinetic friction. We also know Fkfriction = muKFn = muK*mg (Taking Fn = Fg = mg from the vertical force summation).

The horizontal then becomes: ma = Fapplied - muK*mg and rearranging for muK we get

muK = (Fapplied - ma)/ mg

Now all thats left to do is plug in your givens. I ended up with 1.35

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6 years ago

Answered By Navpreet S

Draw the Free body diagram and show all the forces acting on the body. The forces acting here are F= 455N which causes acceleration , friction(f) , gravity force(mg) and the normal reaction( always perpendicular to the contact surface).

Since, the body is not at the rest, it is accelerating in some diection due to force F, we can write the following equation:

F-f = ma, Here f is the frictional force, f=uN, u being the coefficient of kinetic friction

Since there is no motion in vertical direction mg=N

F-umg = ma

u = (F-ma)/mg

u = (455-27*3.6)/(27*9.8) =1.35

 

It is interesting to note here that value of coefficient of kinetic friction is more than one. Usually its in the range[0,1]. However there are few exceptions where the value of friction coefficient can be greater than one.

 Example: Silicone rubber or acrylic rubber-coated surfaces have a coefficient of friction that can be substantially larger than 1.

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