Finding the final speed of a space probe using work and kinetic energy

[aqryus]

Homework Statement:

A deep space probe of mass 4.55 x10^4kg is travelling at an initial speed of 1.22 x10^4
m/s. The engines of the probe exert a force of magnitude 3.85 x10^5
N over 2.45 x10^8 m. Determine the probes final speed

Relevant Equations:

W=Ek2-Ek1
W=fd
W=(1/2)mv2^2-(1/2)mv1^2

First I found work:

W=(3.85x10^5)(2.45x10^8)
W= 9.43x10^13

Then used that for difference of kinetic energy:

9.43x10^13 = (1/2) (4.55x10^4)v2^2 - (1/2)(4.55x10^4)(1.22x10^4)^2
9.43x10^13 = (22750)v2^2 - 3.386x10^12
9.43x10^13 + 3.386x10^12 = (22750)v2^2
9.77x10^13 = 22750v2^2
9.77x10^13/22750 = v^2
squareroot 4.3x10^6 = v
6.55x10^4 = v

But the answer is 1.38x10^4 m/s? I'm not sure where I went wrong. Any help is appreciated

 

[Frabjous]

Are the exponents for the distance and force correct?

 

[aqryus]

Is the exponent for the distance or force correct?

definitely an 8 on the worksheet

 

[Frabjous]

If the work is a factor of 100 smaller, you get their answer. Check the units also. Also check the force.

 

[kuruman]

squareroot 4.3x10^6 = v
6.55x10^4 = v

I don't know about the calculations that brought you to the two last equations but the very last equation does not follow from the next to last equation. The square root of 4.3 is a bit more than 2 and the square root of 10^6 is 10^3. So the answer that follows should have been a bit more more than 2000. You need to redo the calculations.

 

[haruspex]

I don't know about the calculations that brought you to the two last equations but the very last equation does not follow from the next to last equation. The square root of 4.3 is a bit more than 2 and the square root of 10^6 is 10^3. So the answer that follows should have been a bit more more than 2000. You need to redo the calculations.

The 10⁶ is a typo. Should be 10⁹.

 

[Redbelly98]

Something is not adding up here.

The 9.4x10¹³ J of work is roughly 30 times greater than the initial kinetic energy of 0.34x10¹³ J. So, we'd expect the final speed to be a factor of roughly sqrt(30) times the initial speed, as the OP got.

However, the quoted "correct answer" is only about 13% greater than the initial speed (1.38/1.22=1.13). Even if the probe started from rest, doing that much work would give a much greater final speed.

Either there's a typo in the given quantities, or whoever is providing the "correct answer" made an error.

 

[haruspex]

Something is not adding up here.

The 9.4x10¹³ J of work is roughly 30 times greater than the initial kinetic energy of 0.34x10¹³ J. So, we'd expect the final speed to be a factor of roughly sqrt(30) times the initial speed, as the OP got.

However, the quoted "correct answer" is only about 13% greater than the initial speed (1.38/1.22=1.13). Even if the probe started from rest, doing that much work would give a much greater final speed.

Either there's a typo in the given quantities, or whoever is providing the "correct answer" made an error.

As noted by @Frabjous in post #4, taking either the thrust or the distance over which it is applied down two orders of magnitude gives the book answer.