A crate with a mass of 80 kg glides through a space station with a speed of 2.5 m/s. An astronaut speeds it up by pushing on it from behind with a force of 210 N, continually pushing with this force through a distance of 8 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 190 N, backing up through a distance of 6 m. After these two maneuvers, what is the speed of the crate?

To find the final speed of the crate, we need to consider the forces and distances involved in the two maneuvers.

First, let's analyze the acceleration caused by the astronaut pushing the crate from behind. The force exerted by the astronaut is 210 N. Since force is equal to mass multiplied by acceleration (F = ma), we can rearrange this equation to solve for acceleration:
acceleration = force / mass = 210 N / 80 kg ≈ 2.625 m/s^2.

Next, we need to find the change in velocity during the first maneuver. Using the equation:
Δv = a * d,
where Δv is the change in velocity, a is the acceleration, and d is the distance, we can calculate the change in velocity:
Δv = 2.625 m/s^2 * 8 m = 21 m/s.

The initial velocity of the crate is 2.5 m/s, so the final velocity after the astronaut pushes it from behind will be:
final velocity = initial velocity + Δv = 2.5 m/s + 21 m/s = 23.5 m/s.

Moving on to the second maneuver, the astronaut pushes the crate backwards with a force of 190 N. Using the same process as before, we can find the acceleration:
acceleration = force / mass = 190 N / 80 kg ≈ 2.375 m/s^2.

Now, we need to calculate the change in velocity during the second maneuver:
Δv = a * d = 2.375 m/s^2 * 6 m = 14.25 m/s.

Since the crate was initially moving forward at 23.5 m/s and now it's being slowed down, we need to subtract the change in velocity:
final velocity = initial velocity - Δv = 23.5 m/s - 14.25 m/s = 9.25 m/s.

Therefore, the final speed of the crate after the two maneuvers is approximately 9.25 m/s.