Similar question still making a mistake?

Question

A 4.50X10^4 kg space probe is traveling at a speed of 12000 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 3.00X10^5 N over a distance of 2500 km. What is the final speed of the probe?

set this problem up using the formal FD=work and the wnc=final energy-initial energy
750000000=1/2(4.5X10^4)(v-1200)^2-(4.50*10^4)(9.80)
i got 1382.628<<

Answer 1

Using energy:
.5*m*vi^2-F*D=.5*m*vf^2
simplify
vi^2-vf^2=2*F*D/m
vf=sqrt(vi^2-2*F*D/m)
plug in the numbers
vf=sqrt(12000^2-2*300000*2500000/45000)
vf=10519.82 m/s

j

Answer 2

Well, you've got 1200 for the speed in one place and 12,000 in the other. But the main problem is using (v2-v1)^2 where you should have v2^2 - v1^2. Also, what's with the 9.80? Is that meant to be gravitational acceleration? Nothing is stated in g's here, so it's out of place. The equation you need to set up is m/2(v1^1 - v2^2) = FD, and solve for v2.

Answer 3

Ok here's what i found....

We know that:

m = 4.5 x 10^4kg
u (initial speed) = 12000m/s
F = - 3 x 10^5 N
s(distance) = 2500000m
v (final speed) = ?

Using F = ma we can deduce the deceleration to produce this force:

a = F/m
a = -3 x 10^5/4.5 x 10^4
a = -6.67m/s2

Then using the motion equation of v^2 = u^2 + 2as, we can solve for the final speed:

v = (u^2 + 2as)^0.5

v = (12000^2 + 2 x -6.67 x 2500000)^0.5

v = 10520 m/s (nearest whole number)

An alternative to work method i suppose... but same answer :)