maximum static friction acting on the object will be
[tex] F_s = \mu_s mg[/tex]
plug in all values
[tex]F_s = 0.40 \times 15 \times 9.8 = 58.8 N[/tex]
So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force
So here it is given that applied force is 20 N
so here object will not move due to this force and it will remain at rest always
due to this applied force
Answer:
The applied force is [tex]20\;\rm{N,[/tex] then the object will not move due to this force and it will remain at rest always due to this applied force.
Explanation:
Given: A stationary [tex]15\;\rm{kg[/tex] object is located in a table near the surface of the earth. The coefficient of static friction between the surfaces is [tex]0.40[/tex] and the coefficient of kinetic friction is [tex]0.25[/tex].
Weight of the object is [tex]15\;\rm{kg[/tex].
The maximum static friction acting on the object will be [tex]F_s=\mu_s{mg[/tex]
Substiuting all the values:
[tex]F_s=0.40\times15\times9.8\\F_s=6\times9.8\\F_s=58.8\;\rm{N[/tex]
Here, force of static friction is [tex]58.8\;\rm{N[/tex]. it means that if applied force is less than or equal to [tex]58.8\;\rm{N[/tex], then the object will remain stationary as friction can balance the external force upto this limit of external force.
Therefore, the applied force is [tex]20\;\rm{N,[/tex] then the object will not move due to this force and it will remain at rest always due to this applied force.
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