Work and Energy Theorem and Kinetic Energy

[rmarkatos]

A 5.0x10^4kg space probe is traveling at a speed of 11,000m/s through deep space. Retrorockets are fired along the line of motion to reduce the probe's speed. The retrorockets generate a force of 4.0x10^5N over a distance of 2500km. What is the final speed of the probe.

For this problem i know exactly what to do but i know i am getting the wrong answer because I have odd answer selections.

I used W=Fd=1/2m(vf^2)-1/2m(vi^2). I know this is correct because when i plug in the correct answer for vf=9,000 m/s it equals the other side of the equation so i know i am right but for some reason in the process of solving i am not sure what i am doing wrong because i keep getting the wrong answer for vf. It is kind of strange because my units in every part of solving the equation come out correctly.
First i converted km to meters because we work in the mks system

(4x10^5N)(2.5X10^6m)=w
=1x10^12J

1x10^12J=1/2(5x10^4kg)(vf^2)-1/2(5x10^4kg)(11,000m/s)^2

1x10^12J=1/2(5x10^4kg)(vf^2)-3.025x10^12J

from here i tried different ways to solve the equation but each way i still get the wrong answer. I added the 3.025J to both sides and then multiplied the 1/2(5x10^4). Then i divided the 4.025J by the 25,000kg and then solved for vf by taking the square root of both sides but for some reason my answer does not really come out relativlely close 9000m/s

Any suggestions?

 

[Noein]

The work done by the rockets is negative since their force opposes the motion.

 

[rmarkatos]

i am just having a problem solving it algebraically correctly for some reason.

 

[Office_Shredder]

Because your work should be negative. The final kinetic energy - the initial kinetic energy should be negative, yet you have set it to a positive value, meaning v final is going to be greater than v initial. Try making that LHS negative, and it should work.