anonymous
  • anonymous
Work done on this block.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
How much work is need to move a block that is 4.00Kg over a 0.025m distance? |dw:1377888021808:dw| If the force required to over come static friction is 7.84N coefficient of static friction is: 0.2 coefficient of kinetic friction = 0.1 4.00Kg x 9.8 = 39.2 x 0.2 =7.84N 39.2 x 0.1 = 3.92N So the amount of force needed to move a block in that direction is 8N would be enough? Work needed: 0.2J? Is my calculation right here? + Is it even possible to have static friction = kinetic friction?
anonymous
  • anonymous
@Jemurray3 @Vincent-Lyon.Fr @TuringTest @ghazi @Shane_B
anonymous
  • anonymous
@experimentX @.Sam. @amistre64 @ash2326

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anonymous
  • anonymous
Hi @experimentX So is my calculation and my solution correct?
experimentX
  • experimentX
don't know how to work with two friction coefficients. from where to where (up to what velocity) do you consider static friction and from where to where (velocity) do you consider kinetic friction?
anonymous
  • anonymous
Hmm, Im not sure myself.
anonymous
  • anonymous
I mean, static friction is that friction at the beginning...
anonymous
  • anonymous
K.friction is the one that starts as soon as the object starts to move.
anonymous
  • anonymous
I assume, I should be only work on K.friction
anonymous
  • anonymous
But generally, how would you solve this problem? How much work do you need(minimum) to start moving a block of 4kg 0.025m?
experimentX
  • experimentX
i would use Kinetic friction. No work is done with static friction since the body is not moving.
anonymous
  • anonymous
So the force needed to move this object must be greater than 3.92 So, work = 4N x 0.025m = 0.1 J?
anonymous
  • anonymous
Hi @theEric What do you think?
theEric
  • theEric
I just got to the bottom. I'm going to look at the question again and reply! :)
experimentX
  • experimentX
Assuming the body possess velocity from beginning and constant force is being applied to keep it in uniform motion. \[ 4*9.8*0.1*0.025 \approx 0.1 J \]
theEric
  • theEric
So, I would agree that you only need kinetic friction. Work would be \(W=F\ d\ \cos(\theta)\) since the force used to move the box is directed in the same direction as the distance. The force to overcome static friction is applied for no distance, really... So, the force you use to push is opposite to friction. So \(W=-F_{f_\text k}\ d\), and \(F_{f_\text k}=m\ g\ \mu_\text k\).
theEric
  • theEric
I'm assuming it has constant velocity too :)
theEric
  • theEric
That is the minimum work that can be done.
anonymous
  • anonymous
Yes it would have a constant force. So I would require at least 0.1J to move that block correct :D ?
anonymous
  • anonymous
Thanks!
experimentX
  • experimentX
yes
theEric
  • theEric
You're welcome! :)
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
I agree with the 0.1 J derived by experimentX

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