Question

The wind pushes a cup along the sand at the beach. The force of the wind was 17.8N, and the opposing force was 2N. If the cup accelerates at 7m/s^2, how much does the cup weigh?

Answer 1

Answer: 22.05 N

Step-by-step 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:

`F_(x)=F_(1)+F_(2)` (1)

Where:

`F_(1)=17.8N` is the force of the wind

`F_(2)=-2N` is the opposing force (that is why it has the negative sign)

So:

`F_(x)=17.8N-2N=15.8N` (2)

In addition we know by Newton's 2nd law:

`F_(x)=m.a` (3)

Where:

`m` is the mass of the cup

`a=7m/s^(2)` is its acceleration

Substituting (2) in (3):

`15.8N=m(7m/s^(2))` (4)

Finding
`m`:

`m=(15.8N)/(7m/s^(2))=2.25kg` (5)

On the other hand we have the forces acting on the Y axis:

`F_(y)=N-W=0` (6)

Where:

`N` is the normal force

`W=m.g` is the weight of the cup, being
`g=9.8m/s^(2)` the acceleration due gravity

This means:

`N=W=m.g` (7)

Substituting (5) in (7):

`W=(2.25kg)(9.8m/s^(2))` (8)

Finally:

`W=22.05N` This is the weight of the cup