anonymous
  • anonymous
A force of 40N is re
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Sorry the question is: A force of 40N is required to start a 6.0kg box moving across a horizontal concrete dfloor. If 40N force continues the box acceleratesat .90m/s^2. What is the coefficient of kinetic friction?
anonymous
  • anonymous
I already got the coefficient of static friction = .68
anonymous
  • anonymous
^^ Sorry the acceleration should be .80 m/s^2

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anonymous
  • anonymous
I thought the answe would be 4.8 since F/m = a and if m is 6.0kg and acceleration is .80m/s^2, plugging it in would give me 4.8N, but solving for a got me .8 instead.
stormfire1
  • stormfire1
Applied force F = 40N Uk = coefficientof kinetic friction Force of Friction Ff = Uk * mg Fnet = F - Ff = ma \[40N−Uk (9.8m/s^2 ∗6.0kg)N=6.0kg∗0.80m/s^2 =4.8N\]\[40N−4.8N=U_{k}(9.8m/s^2 ∗6.0kg)\]\[35.2N=U_{k}(9.8m/s^2 ∗6.0kg)\]\[\frac{35.2N}{9.8 m/s^2 * 6.0 kg}=0.5986... \]Correcting for significant figures leaves you with a coefficient of friction of ~0.60
anonymous
  • anonymous
That last part of your answer looked something like this: 35.2N/58.8N=.5986?
stormfire1
  • stormfire1
Yes :)
anonymous
  • anonymous
Okay thanks.

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