Force of friction on different times

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SUMMARY

The discussion centers on calculating the force of friction acting on a 3.280 kg block of wood resting on a steel desk, with static friction coefficient μs = 0.605 and kinetic friction coefficient μk = 0.155. At t=0, when a horizontal force F of 12.0 N is applied, the static friction force equals the applied force, resulting in 12.0 N. When the force increases to 24.1 N, the static friction force remains at its maximum of 19.8 N (μs * N), and the block begins to move, transitioning to kinetic friction with a force of 3.9 N (μk * N).

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of static and kinetic friction coefficients (μs and μk)
  • Ability to calculate normal force (N = mg)
  • Familiarity with the concept of forces acting on an object at rest and in motion
NEXT STEPS
  • Calculate the normal force for different weights using N = mg
  • Explore the implications of static versus kinetic friction in real-world scenarios
  • Learn about frictional force calculations in different materials and surfaces
  • Investigate the effects of varying applied forces on motion and friction
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and friction, as well as educators looking for practical examples of force and motion principles.

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Homework Statement



A 3.280 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.605 and the coefficient of kinetic friction is μk = 0.155. At time t = 0, a force F = 12.0 N is applied horizontally to the block. State the force of friction applied to the block by the table at t=0 and t>0:

Consider the same situation, but this time the external force F is 24.1 N. Again state the force of friction acting on the block at t=0 and t>0:

Homework Equations


F = ma
F ≤ μsN
F = μkN
w=mg

The Attempt at a Solution


I tried to find the normal force with mg so I could then multiply this with μs and μk. I thought that the first one would be when time equals 0 but I don't really understand how to use the F = 12(first question) or F = 24.1(second question).
 
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well so what is the static frictional force acting on the block?

What happens to the block if the force that is applied to it is *less than* the static frictional force? What about if the force applied to it is *greater than* the static frictional force?
 

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