# Kinetic Energy Help

Discussion in 'Physics' started by jared11378, Oct 29, 2009.

1. ### jared11378 Thread Starter Member

Oct 29, 2008
11
0
A 5 * 10^4 kg space probe is traveling at a speed of 11,000 m/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 * 10^5 N over a distance of 2,500 km. What is the final speed of the probe?
a. 10 000 m/s
b. 8000 m/s
c. 6000 m/s
d. 9000 m/s
e. 7000 m/s

I used the equation Wnet = 1/2mv^2(final) - 1/2mv^2(initial)
Wnet = Fnet d
Fnet = ma
Probe = (50000)(11000) = 5.5 * 10^8N
retrorockets = 4.0 * 10^5N

Wnet = (5.5 * 10^8N - 4.0 * 10^5)(2500km)
Wnet = 1.374 * 10^12N

Solving for V(final) I get

2(Wnet + 1/2mv^2initial) / m = V(final)^2

2(1.374 * 10^12 + 3.025 * 10^12) / 5 * 10^4 = Vfinal^2

I get 13000 m/s and its not a valid answer of course.

Does anyone see whats going wrong. I was told to use this equation to solve it.

2. ### steveb Senior Member

Jul 3, 2008
2,432
469
You plugged in velecity for accelleration. Also you didn't need to bring in F=ma to try and calculate Fnet. Fnet is given to you in the problem.

3. ### jared11378 Thread Starter Member

Oct 29, 2008
11
0
OK I messed up on a few things hear but how does this look.
d = 2500km = 2.5*10^6m I wasnt converting this to meters

Wnet = 1/2mv^2(final) - 1/2mv^2(initial)
Wnet = -Fnet d

Wnet = (4.0 * 10^5)(2.5*10^6m)
Wnet = -1.0 * 10^12N

Solving for V(final) I get

2(Wnet + 1/2mv^2initial) / m = V(final)^2

2( -1.0 * 10^12 + 3.025 * 10^12) / 5 * 10^4 = Vfinal^2

I get 9000 m/s and it looks correct.

I was thinking Wnet was the vector sum of all forces acting on the object.
This stuff isnt exactly easy reading.