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

In summary: Unfortunately, I'm out of time.In summary, an object with a mass of 161kg moved in outer space and had a speed of 18m/s when it was located at <7,-17,-9>m. A single constant force of <143,308,-396>N acted on the object as it moved to a location <14,-25,-15>m. Using the equations F=ma and v^2=u^2+2as, the final velocity of the object was calculated to be <168.217, 146.696, 176.758>m/s. However, the calculation for the kinetic energy
  • #1
Westin
87
0

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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  • #2
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.
 
  • #3
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 :/

 
  • #4
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.
 
  • #5
does everything else look good besides that?
 
  • #6
You have the correct overall approach.
But note that KE is also a scalar quantity!
 
  • #7
Westin said:

3) Kfinal / mass = Final velocity

Check this.
 

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

1. What is the formula for calculating the final speed of an object after a constant force is applied?

The formula for calculating the final speed of an object after a constant force is applied is v = u + at, where v is the final speed, u is the initial speed, a is the acceleration, and t is the time.

2. How do I calculate the acceleration of an object using the final speed and time?

The acceleration of an object can be calculated using the formula a = (v-u)/t, where a is the acceleration, v is the final speed, u is the initial speed, and t is the time.

3. Is the final speed of an object always the same as the initial speed?

No, the final speed of an object can vary depending on the amount of force applied and the time it takes for the force to act on the object. If the force is applied for a longer period of time, the final speed will be greater than the initial speed.

4. Can the final speed of an object be negative?

Yes, the final speed of an object can be negative if the initial speed is greater than the final speed, meaning the object is slowing down. This can happen if the force applied is in the opposite direction of the initial motion of the object.

5. How does the final speed of an object change if the force applied is doubled?

If the force applied is doubled, the final speed of the object will also double, assuming all other factors remain constant. This is because the acceleration of the object is directly proportional to the force applied, according to Newton's Second Law of Motion.

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