Question

# A bicyclist travels in a circle of radius $$25.0$$ m at a constant speed of $$9.00$$ m/s. The bicyclerider mass is $$85.0$$ kg. Calculatethe magnitudes of (a) the force of friction on the bicycle from theroad and (b) the net force on the bicycle from the road.

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#### The magnitude of the acceleration of the cyclist as he moves along the horizontal the circular path is given by $$v^2/R$$, where v is the speed of the cyclist and R is the radius of thecurve. The horizontal component of Newton’s second law is $$f-s=\dfrac{mv^2}{R}$$where fs is the staticfriction exerted horizontally by the ground on the tires. Similarly, if $$F_N$$ is the vertical the force of the ground on the bicycle and m is the mass of the bicycle and rider, the verticalcomponent of Newton’s second law leads to $$F_N=mg=833N$$. (a) The frictional force is $$f_s=\dfrac{mv^2}{R}=\dfrac{(85.0 kg) (9.00 m/s)^2}{25.0 m}=275 N$$ (b) Since the frictional force $$\vec f_s$$ and $$\vec F_N$$, the normal force exerted by the road, isperpendicular to each other, the magnitude of the force exerted by the ground on thebicycle is therefore $$F=\sqrt{{f_s}^2+{F_N}^2}=\sqrt{(275 N)^2 +(833 N)^2}=877 N$$

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