Question

If homeboy joe is pushing a box with force of 57.0 in the horizontal direction and -15.0 in the vertical direction. What is the magnitude of joes net force?

Answer 1

The magnitude of joes net force is 58.9.

Step-by-step explanation:

The magnitude of the net force is given by:

`F_(net) = \sqrt{F_(x)^(2) + F_(y)^(2)}`

Where:

`F_(x)`: is the force in the horizontal direction = 57.0

`F_(y)`: is the force in the vertical direction = -15.0

`F_(net) = \sqrt{(57.0)^(2) + (-15.0)^(2)} = 58.9`

Therefore, the magnitude of joes net force is 58.9.

I hope it helps you!