Answer:
D. It is no smaller than [tex]0.78N[/tex]
Explanation:
The question is incomplete. These are all the options :
A. It is no larger than [tex]0.78N[/tex]
B. It is [tex]0.78N[/tex]
C. It is [tex]0.08N[/tex]
D. It is no smaller than [tex]0.78N[/tex]
To solve this problem, we have the data of :
[tex]m=0.4kg[/tex]
Where ''m'' is the mass of the block
μ = 0.2
Where ''μ'' is the coefficient of static friction
If we want to find the magnitude of the force of static friction we need to use the following equation :
[tex]F_{sf}=[/tex] μ.[tex]F_{N}[/tex] (I)
Where ''[tex]F_{N}[/tex]'' is the normal force that the desk exerts on the block. Its magnitude is equal to the weight (because we suppose that the block rests horizontally on the desk).
The weight ''[tex]W[/tex]'' can be calculated as :
[tex]W=m.g[/tex]
Where ''m'' is the mass and ''g'' is the acceleration due to gravity.
The value of ''g'' is
[tex]g=9.81\frac{m}{s^{2}}[/tex]
The weight of the block is
[tex]W=(0.4kg).(9.81\frac{m}{s^{2}})[/tex]
[tex]W=3.924N[/tex]
Now, the weight is equal to the normal force ⇒
[tex]W=F_{N}=3.924N[/tex]
Using the equation (I) :
[tex]F_{sf}=(0.2).(3.924N)[/tex]
[tex]F_{sf}=0.7848N[/tex]
The correct option is D. It is no smaller than [tex]0.78N[/tex]