A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following is the best approximation of the box’s acceleration?
120 m/s2
20 m/s2
1 m/s2
9 m/s2

Respuesta :

Answer:

1 m/[tex]s^{2}[/tex]

Explanation:

The Net Force refers to the sum of all forces acting on an object. Hence,

Net Force of box

= 20 + (-9.0)       [It is -9.0 as both forces are in different directions]

= 11.0 N

We also need to know that the net force on an object is equal to the product of the mass of the object and the acceleration of the object. Hence,

[tex]F_{net} = ma[/tex]

11.0 = 12.0[tex]a[/tex]

[tex]a[/tex] = 1 m/[tex]s^{2}[/tex] (nearest whole number)