Calculate Work and Crate Speed in Astronaut's Space Station Maneuvers

[WolfpackPride]

Homework Statement

A crate with a mass of 120 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 250 N, continually pushing with this force through a distance of 5 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 220 N, backing up through a distance of 4 m.

What is the total amount of work done by the astronaut in these two maneuvers? (I correctly calculated this to be W = 370J)


After these two maneuvers, what is the speed of the crate?

Homework Equations

Kf - Ki = work

The Attempt at a Solution

I found the work, but need the speed. I just can't figure out how to pull the velocity out of that correctly

 

[LowlyPion]

I found the work, but need the speed. I just can't figure out how to pull the velocity out of that correctly

If you want speed then what was the KE of the object initially?

How much net work did the astroworker do? If you add that to the object's energy, that's the final KE and Ke = ½mv²

 

[nlightner]

calculated work.

I would like to provide a response to this question by stating that the speed of the crate after the two maneuvers can be calculated using the principle of conservation of energy. The initial kinetic energy of the crate before the astronaut's maneuvers is equal to the final kinetic energy of the crate after the maneuvers. Therefore, we can use the equation Kf - Ki = work to find the final kinetic energy of the crate. We know that the work done by the astronaut is 370J, and the initial kinetic energy (Ki) is equal to 1/2 * m * v^2, where m is the mass of the crate and v is its initial speed. We can rearrange this equation to solve for v, which gives us the final speed of the crate after the maneuvers. Therefore, the speed of the crate after the maneuvers is v = √(2 * work / m) = √(2 * 370J / 120kg) = 3.05 m/s. This means that the astronaut's maneuvers have increased the speed of the crate from 2.5 m/s to 3.05 m/s.