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Normal Force, Static and Kinectic Friction.

1.
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
 

G01

Homework Helper
Gold Member
2,649
16
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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