A 84-kg fireman slides down a 4-m pole by applying a frictional force of 416 N against the pole with his hands. If he slides from rest, how fast is he moving once he reaches the ground?
Fn = mg - Fk,
Fn=84*9.8 - 416 = 407.2 N.=Net force.
Fn = ma,
a = Fn/m = 407.2 / 84 = 4.85 m/s^2.
Vf^2 = Vo^2 + 2a*d,
Vf^2 = 0 + 9.7*4 = 38.78,
Vf = 6.23 m/s.
To determine how fast the fireman is moving once he reaches the ground, we can apply the principles of work and energy.
The work done on an object is equal to the change in its kinetic energy. In this case, the initial kinetic energy is zero, as the fireman starts from rest. The final kinetic energy is given by (1/2)mv^2, where m is the mass of the fireman and v is his final velocity.
The work done on the fireman is equal to the force applied multiplied by the distance over which the force is exerted. The distance traveled by the fireman is the same as the height of the pole, which is 4 meters.
So, the work done by the frictional force can be calculated as follows:
Work = Force x Distance
Work = 416 N x 4 m
Work = 1664 N·m
Since the work done is equal to the change in kinetic energy, we have:
1664 N·m = (1/2)(84 kg)v^2
Simplifying the equation, we can solve for v:
v^2 = (1664 N·m) x 2 / 84 kg
v^2 = 39.619 N·m / kg
v^2 = 39.619 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 6.29 m/s
Therefore, the fireman is moving approximately 6.29 m/s once he reaches the ground.