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

Answer 2

The magnitude of acceleration of the container of the given mass is 1 m/s².

The given parameters:

• Applied, F = 120 N
• Mass of the container, m = 40 kg
• Frictional force, Ff = 80 N

The magnitude of acceleration of the container is calculated by applying Newton's second law of motion as follows;

$\Sigma F = ma\\\\F - F_f = ma\\\\a = \frac{F- F_f}{m} \\\\a = \frac{120 - 80}{40} \\\\a = 1 \ m/s^2$

Thus, the magnitude of acceleration of the container of the given mass is 1 m/s².

Learn more about Newton's second law of motion here: https://brainly.com/question/25545050