What is the Force of Kinetic Friction?

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SUMMARY

The force of kinetic friction for a 10 kg box moving on a level floor with a coefficient of kinetic friction of 0.3 is calculated using the formula fk = FN * μk. The normal force (FN) is determined by the weight of the box, which is FN = mg, resulting in FN = 98 N. Consequently, the force of kinetic friction is calculated as fk = 98 N * 0.3, yielding a final value of fk = 29 N. This discussion emphasizes the importance of understanding the relationship between normal force and friction in physics.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of force, mass, and acceleration
  • Knowledge of the coefficient of friction and its application
  • Ability to perform basic calculations involving multiplication and unit conversions
NEXT STEPS
  • Study the principles of static vs. kinetic friction
  • Learn about the effects of surface materials on friction coefficients
  • Explore advanced topics in dynamics, such as friction in non-horizontal surfaces
  • Investigate real-world applications of friction in engineering and design
USEFUL FOR

Students studying physics, particularly those tackling mechanics and friction concepts, as well as educators seeking to clarify the principles of force and motion.

JustynSC
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Homework Statement


a 10 kg box is moving on a level floor. The coefficient of static friction between the box and floor is 0.3. What is the force of kinetic friction?

Homework Equations


fk = FN * μk

The Attempt at a Solution


I rearranged the formula so that everything was set equal to the normal force, but I'm convinced that there is information missing to solving this.
 
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Looks to me that everything you need is in the problem statement. Use the formula that you quoted to calculate the friction force. What is the normal force in this situation?
 
There is no normal force given. I have no knowledge on how many Newtons are being acted on the 10kg box. Is it required to have an FN given to solve that equation? Otherwise I only have one known variable, the 0.3 constant for static friction. I solved that the FN required to overcome static friction must be: FN>49N. But that is a separate problem that does not state that same force was used in this question. Besides if the the box were pushed at 98N would that change the Fk?
 
JustynSC said:
There is no normal force given. I have no knowledge on how many Newtons are being acted on the 10kg box.
The normal force is easy to calculate from the given information. Hint: What forces act on the box?
 
At rest the box is acted on by Earth's gravity (9.8 m/s2) multiplied by the mass of the box (10kg)and lastly the static friction constant between the two surfaces (0.3) . So the Normal force is in the perpendicular to the surface the box is on. FN= 49 Newtons. But that is fs max, a threshold that once overcome will cause the box to move. So an additional force pushing parallel to the horizontal surface could be as little as 49.00000000001N or les as long as it is greater than 49. Otherwise the box will remain at rest?
 
JustynSC said:
At rest the box is acted on by Earth's gravity (9.8 m/s2) multiplied by the mass of the box (10kg)
Right! That's a downward force acting on the box. The normal force is the upward force from the surface that supports the box. What must that normal force equal?

(Forget all the other stuff for the moment. Just get this problem solved.)
 
(FN=ma=10kg * 9.8m/s2)
So then, FN =98N ?

P.S I am taking physics in college, but have had to teach myself because the professor does not do a great job at explain the reasoning behind the concepts. And his English is pretty limited, so there is a severe language barrier. So, sorry if I am missing something big here, or something obvious for that matter.
 
JustynSC said:
(FN=ma=10kg * 9.8m/s2)
So then, FN =98N ?
You got it.

The normal force must equal the weight of the box (otherwise the box would have a vertical acceleration). Thus FN = mg.

Now just apply the formula you already know for kinetic friction and you're done.
 
Okay just to check my answer:
fk=FN * μk
Therefore: fk=98 N * 0.3μ
Answer: fk= 29 N
 
  • #10
Looks good to me!
 
  • #11
Awesome. Thank you for working that through without just telling me the answer. I think I actually understand the concept of how friction effects an object now.
 
  • #12
JustynSC said:
Awesome. Thank you for working that through without just telling me the answer. I think I actually understand the concept of how friction effects an object now.
Excellent. :thumbup: That's what we try to do here.
 

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