The alarm at a fire station rings and a 87-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.9 m). Just before landing, his speed is 1.3 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Vf^2=Vi^2+2ad where a= 1/mass (mg-friction)

you know Vf, Vi, mg, and distance d.
solve for frictionforce

deg

To find the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole, we can use the following steps:

1. Identify the known quantities:
- Mass of the fireman (m) = 87 kg
- Initial speed (v₀) = 0 m/s (starting from rest)
- Final speed (v) = 1.3 m/s
- Distance traveled (d) = 3.9 m

2. Determine the work done by the frictional force:
The work done by the frictional force can be calculated using the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy.

- The initial kinetic energy (KE₀) = 0 J (starting from rest)
- The final kinetic energy (KE) = 0.5 * mass * velocity^2

Therefore, the work done by the frictional force (W) = KE - KE₀.

3. Find the frictional force (F_friction) using the work done:
Since work (W) = force * distance, we can rearrange the equation as follows:
W = F_friction * d
F_friction = W / d

Substitute the calculated work (W) into the equation and solve for F_friction.

Let's perform the calculations:

- The initial kinetic energy (KE₀) = 0 J
- The final kinetic energy (KE) = 0.5 * mass * velocity^2
KE = 0.5 * 87 kg * (1.3 m/s)^2

- The work done by the frictional force (W) = KE - KE₀

- The frictional force (F_friction) = W / d

Calculate the values step by step and substitute them into the equations to find the magnitude of the frictional force exerted on the fireman as he slides down the pole.

i still cannot solve the problem. can someone please explain it a little bit more?