Force and coefficient of friction problem

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SUMMARY

The discussion centers on calculating the acceleration of a block subjected to an upward force of 0.500mg at a 20-degree angle, with static and kinetic friction coefficients of 0.6 and 0.5, respectively. The user correctly identified the need to apply Newton's second law and set up equations based on a free body diagram. The challenge lies in determining which coefficient of friction to use and eliminating the mass variable from the equations to solve for acceleration. The user also clarified a misunderstanding regarding the unit of mass.

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  • Understanding of Newton's second law of motion
  • Knowledge of free body diagrams
  • Familiarity with static and kinetic friction coefficients
  • Basic algebra for solving equations
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  • Study the application of Newton's second law in frictional scenarios
  • Learn how to differentiate between static and kinetic friction in problem-solving
  • Explore methods for eliminating variables in physics equations
  • Review free body diagram techniques for complex force systems
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Students in physics, educators teaching mechanics, and anyone interested in understanding forces and friction in motion problems.

demonelite123
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A initially stationary block of mass "m" is on the floor. A force of magnitude 0.500mg is then applied at upward angle 20 degrees. What is the magnitude of the acceleration of the block across the floor if coefficient of static friction = 0.6 and coefficient of kinetic friction = 0.5?

i converted 0.5 mg into 5e-7 kg. so i drew a free body diagram and made the following equations: (F_N is the normal force)
F_N + 5e-7(sin 20) = 9.8m
5e-7(cos 20) - Fr = ma.

so i solved for F_N and substituted into second equation:
since Fr = F_N (coefficient of friction) but which coefficient of friction do i use in this scenario? also even if i do find which one to use, i can't eliminate m from the equation so how do i solve for a?
 
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A force of magnitude 0.500*m*g is applied. Here m is the mass of the block.
 
rl.bhat said:
A force of magnitude 0.500*m*g is applied. Here m is the mass of the block.

thank you! i can't believe i mistook mg for milligrams...
 

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