Question

A 5.0 kg box is sliding across a waxed floor by the application of a 15 N east force. If the force of friction is 2.5 N west, what is the net force acting on the box? Whis is the box's acceleration

Answer 1

1) 12.5 N east

There are two forces acting on the box along the horizontal direction:

- The applied force of 15 N east, we indicate it with F

- The force of friction of 2.5 N west, we indicate it with $F_f$

Taking east as positive direction, we can write the two forces has

$F=+15 N\\F_f = -2.5 N$

Therefore, the net force on the box will be:

$F_{net} = F + F_f = 15 + (-2.5) = +12.5 N$

And the positive sign means the direction is east.

2) $2.5 m/s^2$

We can solve this part by using Newton's second law:

$F_{net}=ma$

where

$F_{net}$ is the net force on the box

m is its mass

a is the acceleration

For the box in this problem,

$F_{net} = 12.5 m/s^2$ (east)

m = 5.0 kg

Solving for a, we find the acceleration:

$a=\frac{F_{net}}{m}=\frac{12.5}{5.0}=2.5 m/s^2$

And the direction is the same as the net force (east)