Did I do this physics problem right?

  • Thread starter zeromodz
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In summary, the coefficient of friction value between the box and the floor is 0.2551. This can be calculated by dividing the force applied (25N) by the normal force (mass of box multiplied by gravity) and solving for μ. Since the box is being pushed at a constant velocity, the net force is zero and the acceleration is not needed in the calculation.
  • #1
zeromodz
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Its not for homework, I just want to make sure I can do this.

What is the value of the coefficient of friction between the box and the floor?
A man pushes a box with a mass of 10kg at a constant velocity across the floor. he pushes with a constant force of 25N. What is the coefficient of friction value?


F = ukN
uk = F / N
uk = (25) / ((10)(9.8))
uk = 0.25
 
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  • #2
You will need to know at what rate the box accelerates at in order to find the friction co-efficient.
 
  • #3
LostConjugate said:
You will need to know at what rate the box accelerates at in order to find the friction co-efficient.

Not for this problem. If the velocity is constant, which the problem stated, then the net force is zero.

So mgcosθμ = 25

98μ = 25

μ = .2551...

so yes you're correct.
 

1. Did I use the correct formula for this problem?

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