Question

select the correct answer. elle is pushing a box of mass 7.0 kilograms with the force of 25 newtons. if the force of friction is 2.6 newtons, what is the value of acceleration of the box? a. 0.7 meters/second squared b. 2.6 meters/second squared c. 3.2 meters/second squared d. 3.6 meters/second squared

Explanation:

Step1: Calculate net - force

The net - force $F_{net}$ is the applied force minus the frictional force. So, $F_{net}=F_{applied}-F_{friction}$. Given $F_{applied} = 25\ N$ and $F_{friction}=2.6\ N$, then $F_{net}=25 - 2.6=22.4\ N$.

Step2: Use Newton's second law

Newton's second law is $F = ma$, where $F$ is the net - force, $m$ is the mass, and $a$ is the acceleration. We want to find $a$, so we can re - arrange the formula to $a=\frac{F_{net}}{m}$. Given $m = 7.0\ kg$ and $F_{net}=22.4\ N$, then $a=\frac{22.4}{7.0}=3.2\ m/s^{2}$.

Answer:

C. 3.2 meters/second squared