Constant horizontal pull on a sled

  • #1

Homework Statement


A constant horizontal pull acts on a sled on a horizontal frictionless ice pond. The sled starts from rest. When the pull acts over a distance x, the sled acquires a speed v and a kinetic energy K. If the same pull instead acts over twice this distance, <the sled's speed will be sqrt(2)v and its kinetic energy will be 2K>?


Homework Equations


K=1/2mv^2


The Attempt at a Solution


The last part between the < > was actually the correct answer as part of a multiple choice question, but I've tried figuring out the relationship and haven't come up with anything. I already missed the question (I thought speed would be 2v and kinetic energy would be 2K), but I would still like to know how those figures and values are found for the future. Thanks!
 

Answers and Replies

  • #2
tiny-tim
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hi wondermoose! :smile:

the force is constant, so the acceleration is constant …

so use one of the standard constant acceleration equations to find how v depends on s :wink:
 
  • #3
Hmm... I don't follow. Do I still use the kinetic energy equation in some regard? Or am I trying to complicate the problem?
 
  • #4
ehild
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Homework Statement


A constant horizontal pull acts on a sled on a horizontal frictionless ice pond. The sled starts from rest. When the pull acts over a distance x, the sled acquires a speed v and a kinetic energy K. If the same pull instead acts over twice this distance, <the sled's speed will be sqrt(2)v and its kinetic energy will be 2K>?

Think of the Work-Energy Theorem.

ehild
 
  • #5
tiny-tim
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hi wondermoose! :smile:

(just got up :zzz: …)
Hmm... I don't follow. Do I still use the kinetic energy equation in some regard?

You can either use the usual equations to find v, or, since you're given the distance in this question, you can (as ehild :smile: says) use the work done to find the KE directly …

either will do! :biggrin:
 
  • #6
Alright, so W=Fd
F=ma
W=m*a*d

and a=v/t

so W = m*v*d/t

deltaK=W

1/2mv^2=m*v*d/t

Is that along the right line? I'm so freakin' confused.
 
  • #7
tiny-tim
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hi wondermoose! :smile:

(have a delta: ∆ and try using the X2 icon just above the Reply box :wink:)

why are you introducing a?

∆KE = W = Fd,

so ∆KE is proportional to F, and proportional to d …

carry on from there :smile:
 
  • #8
hi wondermoose! :smile:

(have a delta: ∆ and try using the X2 icon just above the Reply box :wink:)

why are you introducing a?

∆KE = W = Fd,

so ∆KE is proportional to F, and proportional to d …

carry on from there :smile:

But where does velocity fit into those relationships? I know the .5mv2 is going to come into play somewhere along the way, but I don't know where or how. Thanks, tim!
 
  • #9
tiny-tim
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speed is √(KE/2m)
 
  • #10
speed is √(KE/2m)

Alright, so if K=Fd, and F is constant and d doubles, therefore K doubles. Then, relating back to
√(KE/2m), then if K is 2K then the speed (v) would still be √(2/2m).

v=√(2/2m)

Closer? Thanks
 

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