For this case we must draw the free body diagram of the apple.
It is observed in the attached figure that we have the force of the wind acting to the right, in addition to the force of the weight that goes down.
[tex]Mg[/tex]: It is the force exerted by the weight of the apple (M: It is the mass and g: It is the acceleration of gravity)
[tex]F_ {1}:[/tex] It is the force exerted by the wind
Then, the net force will be given by:
[tex]F_ {n} = \sqrt {(Mg) ^ 2 + (F_ {1}) ^ 2}[/tex]
We have to:
[tex]Mg = 0.25kg * 9.8 \frac {m} {s ^ 2} = 2.45N\\F_ {1} = 1.05N[/tex]
Substituting we have:
[tex]F_ {n} = \sqrt {(2.45) ^ 2 + (1.05) ^ 2}\\F_ {n} = \sqrt {6.0025 + 1.1025}\\F_ {n} = \sqrt {7.105}\\F_ {n} = 2.66[/tex]
Thus, the net force acting on the apple is 2.66N
Answer:
Option D
