Answer: 22.05 N
Explanation:
Let's begin with the cup's free body diagram (figure attached), there we are able to see the forces acting on this object:
On the X axis:
[tex]F_{x}=F_{1}+F_{2}[/tex] (1)
Where:
[tex]F_{1}=17.8N[/tex] is the force of the wind
[tex]F_{2}=-2N[/tex] is the opposing force (that is why it has the negative sign)
So:
[tex]F_{x}=17.8N-2N=15.8N[/tex] (2)
In addition we know by Newton's 2nd law:
[tex]F_{x}=m.a[/tex] (3)
Where:
[tex]m[/tex] is the mass of the cup
[tex]a=7m/s^{2}[/tex] is its acceleration
Substituting (2) in (3):
[tex]15.8N=m(7m/s^{2})[/tex] (4)
Finding [tex]m[/tex]:
[tex]m=\frac{15.8N}{7m/s^{2}}=2.25kg[/tex] (5)
On the other hand we have the forces acting on the Y axis:
[tex]F_{y}=N-W=0[/tex] (6)
Where:
[tex]N[/tex] is the normal force
[tex]W=m.g[/tex] is the weight of the cup, being [tex]g=9.8m/s^{2}[/tex] the acceleration due gravity
This means:
[tex]N=W=m.g[/tex] (7)
Substituting (5) in (7):
[tex]W=(2.25kg)(9.8m/s^{2})[/tex] (8)
Finally:
[tex]W=22.05N[/tex] This is the weight of the cup
