Finding speed given force and distance

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Homework Help Overview

The discussion revolves around calculating the final speed of a skier given a constant force, mass, and distance. The problem is situated within the context of classical mechanics, specifically involving concepts of force, work, and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate force, distance, and acceleration but expresses confusion about incorporating time into their calculations. Some participants clarify the relationship between force and mass, and introduce the work-energy theorem as a potential approach. Others question the original poster's understanding of units, particularly regarding speed versus acceleration.

Discussion Status

Participants are exploring different interpretations of the problem and discussing relevant equations. Some guidance has been offered regarding the work-energy theorem and the relationships between work, force, and kinetic energy. There is an ongoing dialogue about the correct units for speed and acceleration, with no explicit consensus reached on the final calculation.

Contextual Notes

There is mention of a potential typo in the problem statement regarding the units for speed, which has led to some confusion among participants. The original poster's assumption about the relationship between force and distance is also under scrutiny.

fattydq
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Starting from rest, a constant force of 45 N acts on a 67 kg skier over a distance of 8.1 m. What is the skier's final speed in m/s^2?

Now I'm assuming since force is distance times acceleration, I can take 45 and divide it by 8.1 to find the acceleration. But even still, then I will have distance and acceleration, but no unit of time to take into account. Any suggestions? This is my last post of the day I swear! : P
 
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Force is NOT distance times acceleration, it is F=ma.

Have you learned the work energy theorem? Remember that the work is force times distance or W=Fd. According to the work energy theorem this is also equal to the change in kinetic energy.
 
206PiruBlood said:
Force is NOT distance times acceleration, it is F=ma.

Have you learned the work energy theorem? Remember that the work is force times distance or W=Fd. According to the work energy theorem this is also equal to the change in kinetic energy.

Right, mass times acceleration, not distance times acceleration, my mistake. I still don't understand how, given the figures I'm given, I can use Work equals force times distance to find SPEED?
 
These are the relationships you need to know to solve this.

W=Fd
[tex]w=\Delta E_k[/tex]
[tex]E_k=\frac{1}{2}mv^2[/tex]

Once you find the work you have simultaneously found the change in kinetic energy. Assuming the skier starts at rest, then the change in kinetic energy is the total kinetic energy. Solve for velocity.
 
206PiruBlood said:
These are the relationships you need to know to solve this.

W=Fd
[tex]w=\Delta E_k[/tex]
[tex]E_k=\frac{1}{2}mv^2[/tex]

Once you find the work you have simultaneously found the change in kinetic energy. Assuming the skier starts at rest, then the change in kinetic energy is the total kinetic energy. Solve for velocity.
So 45 times 8.1=364.5
364.5=\Delta E_k[/tex]
And thus 364.5=1/2 (67) times velocity^2 which turns into
729=67*V^2
729 Divided by 67 =10.88 m/s^2 which is velocity squared, taking the square root of that gives me

3.298575 m/s^2

Is this correct?
 
yep yep :D

Well the units should just be meters per second, yours represents acceleration!
 
206PiruBlood said:
yep yep :D

Well the units should just be meters per second, yours represents acceleration!

Strange...on my worksheet it says "What is the skier's final speed in m/s^2?" So I just assumed those were the units...
 
Well regardless, thanks a lot for the help! : )
 
nah that is wrong i now u need to now this f=m x a so we don't now a so we do it like this f=m x a /m then we go like this to a=f/m and btw mos je i deqanit
 
  • #10
the pro said:
nah that is wrong i now u need to now this f=m x a so we don't now a so we do it like this f=m x a /m then we go like this to a=f/m and btw mos je i deqanit

I'm going to go ahead and ignore your post since half of it isn't even in proper English...no offense...And beside it asks for speed not acceleration...
 
  • #11
Well meters per second squared is an acceleration and meters per second is a velocity. I am assuming its just a typo.
 
  • #12
206PiruBlood said:
Well meters per second squared is an acceleration and meters per second is a velocity. I am assuming its just a typo.

Yeah, me too, I'm assuming my professor just meant m/s. I'll have to bring it up to her in class tomorrow.
 

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