Solving Sliding Friction Homework: F=ma & Nuk

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SUMMARY

This discussion addresses the application of Newton's second law (F=ma) in the context of sliding friction on a horizontal surface. The key equation derived is Fapplied - Ffric = ma, where Fapplied is the force applied to accelerate the object and Ffric is the force of friction opposing the motion. The conversation emphasizes the need for additional information, such as the final speed after the initial force and the distance traveled during deceleration, to solve for the forces involved accurately. The analysis concludes that without knowing the time interval of the applied force, the exact force of the initial hit cannot be determined.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of friction coefficients, specifically kinetic friction (μk)
  • Familiarity with kinematics equations for motion
  • Basic concepts of impulse and momentum
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  • Study the relationship between force, mass, and acceleration in dynamic systems
  • Explore the calculation of frictional forces using the formula Ffric = μk * N
  • Learn about kinematic equations and their application in non-constant acceleration scenarios
  • Investigate impulse and its relation to changes in momentum in physics
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces and motion involving friction.

gcharles_42
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Homework Statement



So say there was some object on a Horizontal surface. id there was some force on (not constant, but a hit or something) this object and there was a sliding friction between the surface and the object. How would that factor into the F=ma formula? and what if you are given a mass, and a distance that the object moved (not acceleration), and sliding friction coefficient of course - how would you find the force?

Homework Equations



F=ma

Nuk (N)= Force of Friction

The Attempt at a Solution



Would it be F - F(of friction) = m (a)?

converting that accel to a distance is what is bothering me a lot :/
 
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Hello,

The F in F = ma is always the NET force acting on the object. So, yes, in the beginning, when a force is briefly applied to accelerate the object, the equation would be:

Fapplied - Ffric = ma

Once the applied force has ceased, and the object slides to a stop under friction, it is just:

-Ffric = ma

The kinematics equations you probably know are only applicable for constant acceleration. In this situation, the acceleration is not constant, because the net force is not constant with time. For a brief period at the beginning, it includes the initial "hit", and afterwards, it includes only friction. So, you need more information in order to figure things out.

If you knew what final speed the object reached at the end of the initial "hit" (the period during which it was accelerating), and you know the distance over which it traveled *after* reaching that max speed, then this reduces to a standard kinematics problem and you can solve for the decelerating frictional force.
 
Hmm, ok, this problem is a little more constrained than I first thought. *IF* I assume that the acceleration period (the period of the "hit") was very brief, then I can assume that most of the distance given in the problem was spent decelerating. In the limit where I assume that the object was decelerating over all that distance, then I know the following:

I know vf = 0

I know the value of 'a' *during the deceleration*, because I know m, which means I know N (it's a horizontal surface), and I know μk, which means I know Ffric, and that gives me a.

I know d

From these three things, I can solve for vi, i.e. how fast must it have been moving just after the "hit" in order for it to require that amount of distance to slow to a stop at that acceleration?

Since I know vi, I know how much change in momentum, or *impulse*, was delivered by the hit. I can't solve for Fapplied (the "hit") from this impulse without knowing over what time interval the hit was delivered. So that's the end of what we can determine.
 

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