The alarm at a fire station rings and a 98.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.00 m). Just before landing, his speed is 1.81 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

PE max = Mg*h-Fk*h = 0.5*M*V^2

M = 98.5 kg
g = 9.8 m/s^2
h = 4 m
V = 1.81 m/s.
Solve for Fk.

Divide both sides by M, and solve for Fk

Oh, sliding firemen, huh? That's a hot topic! So, let's slide into the problem and find the answer.

First, let's calculate the fireman's initial potential energy when he's at the top of the pole. We can use the formula: PE = m * g * h, where m is the mass of the fireman (98.5 kg), g is the acceleration due to gravity (approx. 9.8 m/s^2), and h is the height of the pole (4.00 m).

PE = 98.5 kg * 9.8 m/s^2 * 4.00 m

Next, let's calculate the fireman's final kinetic energy just before landing. We can use the formula: KE = (1/2) * m * v^2, where m is the mass of the fireman (98.5 kg) and v is the final velocity (1.81 m/s).

KE = (1/2) * 98.5 kg * (1.81 m/s)^2

Now, since energy is conserved, the potential energy at the top should be equal to the kinetic energy just before landing. So, we can set PE equal to KE:

PE = KE

98.5 kg * 9.8 m/s^2 * 4.00 m = (1/2) * 98.5 kg * (1.81 m/s)^2

Now, let's solve for the kinetic frictional force. The frictional force is given by the equation: F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

In this case, the frictional force is the force that opposes the motion of the fireman, so it's pointing upward. The normal force is equal to the weight of the fireman.

F_friction = μ * m * g

Now, let's plug the values we have:

F_friction = μ * 98.5 kg * 9.8 m/s^2

And there you have it, the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole.

To find the magnitude of the kinetic frictional force exerted on the fireman, we can use the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the gravitational force is equal to the work done by the frictional force.

The work done by the gravitational force is given by the formula:

Work_gravitational = m * g * h

where:
m = mass of the fireman = 98.5 kg
g = acceleration due to gravity = 9.8 m/s^2
h = vertical height = 4.00 m

Work_gravitational = 98.5 kg * 9.8 m/s^2 * 4.00 m
= 3861.6 J

The work done by the frictional force is given by the formula:

Work_friction = -F_friction * d

where:
F_friction = magnitude of the frictional force (what we want to find)
d = distance traveled = 4.00 m

Work_friction = -F_friction * 4.00 m

Since the fireman starts from rest and has a speed of 1.81 m/s just before landing, his change in kinetic energy is equal to the final kinetic energy:

Change_in_kinetic_energy = 0.5 * m * v_f^2

where:
v_f = final speed = 1.81 m/s

Change_in_kinetic_energy = 0.5 * 98.5 kg * (1.81 m/s)^2
= 160.40565 J

According to the work-energy principle, the work done by the gravitational force is equal to the work done by the frictional force, so:

3861.6 J = -F_friction * 4.00 m + 160.40565 J

Rearranging the equation to solve for the magnitude of the frictional force:

F_friction = (3861.6 J - 160.40565 J) / 4.00 m

F_friction = 3701.19435 J / 4.00 m

F_friction = 925.2986 N

Therefore, the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole is approximately 925.3 N.

To find the magnitude of the kinetic frictional force exerted on the fireman, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration:

Fnet = m * a

In this case, the net force is the sum of the gravitational force (mg) and the frictional force (Ffriction) acting on the fireman. Since the fireman is sliding down the pole, we know that the frictional force is acting in the opposite direction to the motion, which means it is directed upwards.

The gravitational force acting on the fireman can be calculated using the formula:

Fgravity = m * g

where m is the mass of the fireman and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since the fireman is sliding down the pole, he is experiencing an acceleration due to the net force acting on him. Therefore, we can rearrange Newton's second law to solve for the magnitude of the frictional force:

Ffriction = Fnet - Fgravity

To find the net force, we need to determine the fireman's acceleration. We can use the following kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (1.81 m/s), u is the initial velocity (0 m/s since he is starting from rest), a is the acceleration, and s is the distance traveled (4.00 m).

Rearranging the equation to solve for acceleration, we have:

a = (v^2 - u^2) / (2s)

Now, we can substitute the values into the equation:

a = (1.81^2 - 0^2) / (2 * 4.00)

After calculating, we find that a = 0.8225 m/s^2.

Substituting the values of mass (m = 98.5 kg), acceleration (a = 0.8225 m/s^2), and gravitational force (g = 9.8 m/s^2) into the formula for the net force:

Fnet = m * a = 98.5 * 0.8225

After calculating, we find that Fnet = 80.97125 N.

Finally, we can calculate the magnitude of the frictional force:

Ffriction = Fnet - Fgravity = 80.97125 - (98.5 * 9.8)

After calculating, we find that Ffriction = 19.1775 N.

Therefore, the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole is approximately 19.1775 N.