Determine the probeʼs final speed

[jbjohnybaker]

need help!

A deep-space probe of mass 4.55 x 10^4 kg is traveling at an initial speed of 1.22 x 10^4 m/s. The engines of the probe exert a force of magnitude 3.85 x 10^5 N over 2.45 x 10^6 m. Determine the probeʼs final speed. Assume that the decrease in mass of the probe (because of fuel being burned) is negligible.
-------------------
m = 4.55 x 10^4 kg
Vi = 1.22 x 10^4 m/s
Fa = 3.85 x 10^5 N
D= 2.45 x 10^6 m

----------------------
w = (F)(D) (cos 0)
wtotal = 1/2mv^2f - 1/2mv^2vi

------------------
w= (4.55x10^4)(2.45x10^6)
w = 1.11475x 10^11

Wtotal = 1/2mv^2f - 1/2mv^2vi
1.11475x 10^11 = 1/2(4.55x10^4)(vf^2) - 1/2 (4.55x10^4)(1.22x10^4)
1.11475x 10^11 = 1/2(4.55x10^4)(vf^2) - 3.38611 x10^12
1.11475x 10^11 + 3.38611 x10^12 = 1/2 4.55x10^4 (V^2)
v = sqrt ( 2(3.498x1012) / 4.55x104

v= 1.24 x10^4 m/s??
The real answer is 1.38 x10^4 m/s

 

[SteamKing]

Rather than KE, why don't you try that old standby F = ma. Remember, you have an initial velocity specified before F is applied, and you know the total distance traveled during which F is applied.

 

[gneill]

-------------------
m = 4.55 x 10^4 kg
Vi = 1.22 x 10^4 m/s
Fa = 3.85 x 10^5 N
D= 2.45 x 10^6 m

----------------------
w = (F)(D) (cos 0)
wtotal = 1/2mv^2f - 1/2mv^2vi

------------------
w= (4.55x10^4)(2.45x10^6)
w = 1.11475x 10^11

You've multiplied mass by distance rather than force by distance, so you've got an incorrect value for the work done by the engine over the given distance.

 

[jbjohnybaker]

You've multiplied mass by distance rather than force by distance, so you've got an incorrect value for the work done by the engine over the given distance.

I tried
W = (3.85^5)(2.45x10^6)
= 9.43x10^12 J
And I did the rest but still don't the right answer

 

[jbjohnybaker]

Rather than KE, why don't you try that old standby F = ma. Remember, you have an initial velocity specified before F is applied, and you know the total distance traveled during which F is applied.

But i don't know what the acceleration is :S

 

[SammyS]

You know the mass and the force, right?

 

[gneill]

I tried
W = (3.85^5)(2.45x10^6)
= 9.43x10^12 J
And I did the rest but still don't the right answer

You're off by a power of ten. Should be 9.43x10¹¹ J.

 

[jbjohnybaker]

You know the mass and the force, right?

yes the mass and the force are known

 

[jbjohnybaker]

W = (F)(D) cos 0
W = (3.85x10^5)(2.45x10^6) cos 0
W = 9.4 x 10^11 J

W = 1/2mv^2f - 1/2mv^2i
9.4x10^11 = 1/2(4.55^14)vf^2 - 1/2 (4.55x10^4)(1.22x10^4)^2
9.4x10^11 = 1/2(4.55^14)vf^2 - 3.38611 x10^12
9.4x10^11 + 3.38611 x10^12 = 1/2(4.55^14)vf^2
4.8011 x10^12 = 1/2(4.55^14)vf^2

vf = sqrt (2(4.8011x10^12))/ (4.55x10^4)
vf = 1.46x10^4 m/s <-- not the right answer

the correct answer is 1.38x10^4 m/s

 

[SteamKing]

Well Johnny, you've hit the wall again. Sorry. Why don't you try F = ma? It's so easy. You've got F. You've got m. Don't you think you could get a? Once you've got a, it's constant over the entire distance travelled. m doesn't change. You know the initial velocity and the total distance travelled. Don't you think you could find t for me, Johnny? Then you could be happy, Johnny. Once you know a, t, and the initial velocity, everything will be all right. Don't you think so, Johnny?

 

[ehild]

W = (F)(D) cos 0
W = (3.85x10^5)(2.45x10^6) cos 0
W = 9.4 x 10^11 J

Do not round so much during calculations. The data are give with 3 significant digits, keep one more (four ones).

W = 1/2mv^2f - 1/2mv^2i
9.4x10^11 = 1/2(4.55^14)vf^2 - 1/2 (4.55x10^4)(1.22x10^4)^2
9.4x10^11 = 1/2(4.55^14)vf^2 - 3.38611 x10^12

what is that 4.55^14?

You applied the Work Theorem, and it is a best method when the force and displacement are given and the final speed is the question. Calculate it again but do not drop significant digits.

ehild

 

[ehild]

Well Johnny, you've hit the wall again. Sorry. Why don't you try F = ma?

You are not right. It is no need to get the time of acceleration. Johnny applied a much shorter and easier method to get the final speed.

ehild

 

[SteamKing]

He's getting the wrong speed because he messed up his addition in post #9.
4.33*10^12 = 0.5*m*vf^2

And my method works perfectly well. After all, I was able to calculate vf.

 

[jbjohnybaker]

i solved it..thank you very much