A 82 kg runner with a speed of 5.6 m/s makes a sharp turn with a radius of 1.1 m to avoid a collision. What is the magnitude of the frictional force (in N) the ground exerts on the runner during the turn?
mv^2/r = Ff
Plug. Chug.
To find the magnitude of the frictional force the ground exerts on the runner during the turn, we can use Newton's second law and the concept of centripetal force.
1. First, let's find the magnitude of the centripetal force acting on the runner. The centripetal force is the force required to keep an object moving in a curved path. It can be calculated using the formula:
F = m * (v^2 / r)
where F is the centripetal force, m is the mass of the runner (82 kg), v is the speed of the runner (5.6 m/s), and r is the radius of the turn (1.1 m).
Plugging in the given values, we have:
F = 82 kg * (5.6 m/s)^2 / 1.1 m
2. Calculate the centripetal force using the provided values:
F = 82 kg * (31.36 m^2/s^2) / 1.1 m
F = 2353.6 N
3. The frictional force between the runner and the ground provides the centripetal force necessary to keep the runner in the curved path. Therefore, the magnitude of the frictional force is equal to the magnitude of the centripetal force.
So, the magnitude of the frictional force exerted by the ground on the runner during the turn is approximately 2353.6 N.