A person pushes against a 9.2-kg box with a force of 11.5 N. The frictional force on the box is 2.1 N. What is the acceleration of the box? (1 point)

0.86 m/s2
0.92 m/s2
1.0 m/s2
1.3 m/s2

im seriously not getting this lesson :/

Question

A person pushes against a 9.2-kg box with a force of 11.5 N. The frictional force on the box is 2.1 N. What is the acceleration of the box?  (1 point)

0.86 m/s2
0.92 m/s2
1.0 m/s2
1.3 m/s2

im seriously not getting this lesson :/

anonymous · Answer

the box has a forward force of 11.5-2.1= 9.4N
A=F/M = 9.4/9.1 = 1.03 m/s2

I believe that's how you should calculate it... not 100% sure about the last part since I haven't done this in a while...

anonymous · Answer

Ok it seems to make sense to me :)

anonymous · Answer

Newton's Second Law states that the net force acting on an object is equal to the mass times acceleration. $$\sum F = ma$$

Let's draw a free body diagram. |dw:1339007613257:dw|$F_a$ is te applied force from the person, $F_f$ is the frictional force, positive x is to the right. 

$F_a$ is positive since it acts to the right, while $F_f$ is negative since it acts in the opposite direction. 

We can now right Newton's Second Law as$$F_a - F_f = ma$$This can be solved for a with a little algebra.