Question

A 102 kg motorcycle is driving around a turn with a radius of 50.5 m at a constant speed of 12.7 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.

Answer 1

The minimum force of friction required to prevent the motorcycle from sliding off the road is approximately 324 N.

Step-by-step explanation:

To calculate the minimum force of friction required to prevent the motorcycle from sliding off the road, we can consider the forces acting on the motorcycle. In this case, we have the gravitational force (mg) acting downwards and the centripetal force (mv²/r) acting towards the center of the turn. The force of friction (f) opposes the motion and prevents the motorcycle from sliding off.

Since the motorcycle is not accelerating, the net force in the centripetal direction is zero. Therefore, the force of friction is equal to the centripetal force: f = mv²/r. Plugging in the values, we have f = (102 kg)(12.7 m/s)² / 50.5 m. Solving this equation gives us a force of friction of approximately 324 N.