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A skateboarder is trying to take a sharp turn (radius of 1.2m). She grabs hold of a tree branch to help make the turn. She pulls with 45 N of force. If she makes the turn going 3.2 m/s, what is the coefficient of friction between the wheels and ground. The mass of the girl and board together is 35 kg. Identify the F net equation and coefficient of friction value that best fits the above situation.

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Answer:

Explanation:

The skateboarder moves on a circular path of radius 1.2 m with velocity 3.2 m /s so centripetal force required

= m v² / r , m is mass , v is velocity and r is radius of the circular path.

= 35  x 3.2² / 1.2

= 298.67 N

Force 45 N is provided by tree branch . The rest will be provided by frictional force

frictional force F  = 298.67 - 45

= 253.67 N

weight = reaction force of ground = 35 x 9.8

= 343 N

Coefficient of friction = friction / reaction force

= 253.67 / 343

= .74 .