Question

A 131 kg motorcycle is driving around a turn with a radius of 72.9 m at a constant speed of 14.3 m/s. What is the minimum force of friction required to prevent the cycle from sliding off of the road? Report your answer to the nearest whole number. Do not include the units in your answer, or the computer will mark the problem wrong.

A) 764 N
B) 675 N
C) 845 N
D) 928 N

Answer 1

The minimum force of friction required to prevent the motorcycle from sliding off the road is approximately 256 N.

Step-by-step explanation:

To calculate the minimum force of friction required to prevent the motorcycle from sliding off the road, we need to consider the centripetal force acting on the motorcycle. The centripetal force is provided by the frictional force between the motorcycle's tires and the road surface.

Centripetal force = mass x (velocity^2 / radius)

Substituting the given values into the equation, we get:
Centripetal force = 131 kg x (14.3 m/s)^2 / 72.9 m = 256.37 N

Therefore, the minimum force of friction required to prevent the motorcycle from sliding off the road is approximately 256 N, which rounds to 256 N.