A 75kg firefighter slides down a 2.5m fire pole from rest. It takes the firefighter 1.8s to reach the ground. If the firefighter bends his knees when he reaches the ground such that his centre of mass mives 41cm while coming to a stop, what must the normal force of the ground acting on him have been during the stopping phase of his motion?
(I thought if I just found the Force of gravity i could solve for the normal force but I think that's too simple. Basically, not sure how to approach this question.)
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the firefighter is sliding down the fire pole, and we need to find the normal force acting on him when he comes to a stop.
First, let's calculate the acceleration of the firefighter. We can use the kinematic equation:
v = u + at
Where:
v = final velocity = 0 m/s (the firefighter comes to a stop)
u = initial velocity = 0 m/s (the firefighter starts from rest)
t = time = 1.8 s
Rearranging the equation, we have:
a = (v - u) / t
a = (0 - 0) / 1.8
a = 0 m/s²
Since the acceleration is zero, we know that the net force acting on the firefighter must also be zero. This means that the normal force and the force of gravity must balance each other out.
The force of gravity acting on the firefighter can be calculated using:
F = mg
Where:
m = mass = 75 kg
g = acceleration due to gravity = 9.8 m/s²
F = 75 kg * 9.8 m/s²
F = 735 N
Now, when the firefighter bends his knees and comes to a stop, we need to find the normal force exerted by the ground. Since the net force is zero, this normal force will have the same magnitude but opposite direction to the force of gravity.
Therefore, the normal force acting on the firefighter when he comes to a stop is 735 N, directed upward.