Final Speed of Object After Constant Force Applied - 31.358m/s

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An object with a mass of 161 kg in outer space initially moves at 18 m/s when a constant force of <143,308,-396> N is applied. The acceleration calculated from the force is <.888,1.913,-2.46> m/s². Using the kinematic equation v² = u² + 2as, the final velocity components were determined as <18.342,17.192,18.802>. The magnitude of the final speed was computed to be 31.358 m/s, but there were discussions about using work and energy concepts instead. The importance of recognizing work as a scalar quantity and kinetic energy as well was emphasized in the calculations.
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Homework Statement


An object with mass 161kg moved in outer space. When it was at a location [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png7,-17,-9[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngm its speed was 18m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs. A single constant force [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png143,308,-396[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngN acted on the object while it moved to a location [PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char68.png14,-25,-15[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmsy10/alpha/144/char69.pngm. What is the speed of the object at this final location?[/B]

Homework Equations


W=Fd
F=ma
KE=1/2mv^2
v^2=u^2+2as

The Attempt at a Solution



1) Used F=ma
<143,308,-396> = 161a
a=<.888,1.913,-2.46>m/s

2) Plugged numbers into v^2=u^2+2as to get final velocity, I did it 3 times because of vectors so I got Vf= <18.342,17.192,18.802>

3) I took <18.342,17.192,18.802> and did sqrt(18.342^2+17.192^2+18.802^2) and got 31.358m/s[/B]
 
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Hello, Westin.

For the x-component, you would have ##v_x^2 = u_{x}^2 + 2a_x\Delta x##. But, you don't know ##u_x##.

I think work and energy concepts are going to be a better approach here.
 
The section we are covering is currently energy so I think you're right.

Second attempt:
1) W=Fd
W= (<143,308,-396> )(<7,-8,-6>)
W= <1001,-2464,2376>

2) Kfinal = Kintial + W

Kinital = (1/2)mv^2 = (1/2)(161)(18)^2 = 26082J

Kfinal = 26082J + <1001,-2464,2376>J

Kfinal = <27083, 23618, 28458> J

3) Kfinal / mass = Final velocity
<27083, 23618, 28458> / 161 = <168.217, 146.696, 176.758>m/s

4) sqrt(168.217^2+146.696^2+176.758^2) = 284.71m/s

Incorrect once again :/

 
Westin said:
The section we are covering is currently energy so I think you're right.

Second attempt:
1) W=Fd
W= (<143,308,-396> )(<7,-8,-6>)
W= <1001,-2464,2376>

Work is a scalar quantity, not a vector quantity.
 
does everything else look good besides that?
 
You have the correct overall approach.
But note that KE is also a scalar quantity!
 
Westin said:

3) Kfinal / mass = Final velocity

Check this.
 

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