Question

1 point

A box weighing 12 Newtons is dragged across a
rough floor with a constant velocity of 1.4 meters
per second. The horizontal applied force pulling
the box is 8.5 Newtons. What is the magnitude of
the friction force the floor exerts on the box?
Your answer
PLEASE ANSWER

Answer 1

The magnitude of the friction force that the floor exerts on the box is 8.5 newtons.

Step-by-step explanation:

From Newton's Laws of Motion and given that box is moving at constant velocity, the net acceleration of the box is zero, meaning that magnitude of friction force is equal to the magnitude of the horizontal applied force. Please notice that friction force is the reaction to the horizontal force. The equation of equilibrium for the box is:

`\Sigma F = F - f = 0` (1)

Where:

`F` - Horizontal applied force, measured in newtons.

`f` - Friction force, measured in newtons.

If we know that
`F = 8.5\,N`, then the magnitud of the friction force that the floor exerts on the box is:

`f = F`

`f = 8.5\,N`

The magnitude of the friction force that the floor exerts on the box is 8.5 newtons.