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

Answer:

The magnitude of joes net force is 58.9.

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!