Normal Force, Static and Kinectic Friction.

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SUMMARY

The discussion focuses on the application of static and kinetic friction to determine whether a block with a weight of 41.0 N will move when a horizontal force of 48 N is applied. The coefficients of static and kinetic friction are 0.650 and 0.420, respectively. Calculations reveal that the maximum static frictional force (FsMAX) is 26.65 N, indicating that the applied force exceeds this threshold, resulting in movement. The block's acceleration is calculated to be 2.14 m/s² using Newton's Second Law.

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  • Understanding of Newton's Second Law of Motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to perform basic physics calculations involving forces and acceleration
  • Familiarity with the concept of normal force (FN)
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A block whose weight is 41.0 N rests on a horizontal table. A horizontal force of 48 N is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420 N, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? (If not, enter 0.)



2. W (weight) = mg (mass)(9.8m/s^2)= normal force (FN), FsMAX (maximum static frictional foroce) = μs (coefficient of static friction) x FN, Ax (acceleration in x component) = Fk (kinetic frictional force) / m = ( μk x FN) / m. μk = coefficient of kinetic friction

Attempt 1)
3.
Attempt 1)
FsMAX = μs x FN
= (.650 x 41)
= 26.65
48 (Horizontal force) > 26.65 --> acceleration will occur
Ax = Fk/m
= μk x Fn / m
= (.420 x (41 - 26.65) ) / (41 / 9.8)
= (.420 x 14.35) / 4.1836
= 1.44 m/s ^2
----------------------
Attempt 2)
Ax = Fk/m
= μk x Fn / m
= (.420 x (48 - 26.65) ) / (41 / 9.8)
= (.420 x 21.35) / 4.1836
= 2.14 m/s ^2
 
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Remember that the acceleration will be given by Newton's Second Law:

[tex]a=\frac{\Sigma F}{m}[/tex]

What is [tex]\Sigma F[/tex]?
 
Last edited:

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