Question

A force of 120 N is exerted on a 40 kg container which sits on a floor. If the frictional force between floor and container is 80 N, what is the magnitude of the acceleration of the container?

Answer 1

Answer:

Magnitude is 144N .

Explanation:

a²+b²=c²

120N²+80N²=c²

14400+6400=c²

$\sqrt{20800} = \sqrt{c} ^{2}$

144N=c