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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 :

The acceleration of the box is approximately [tex]1 m/s^2[/tex]

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

[tex]\sum F = ma[/tex]

where

[tex]\sum F[/tex] is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

[tex]\sum F = F_a - F_f[/tex]

where

[tex]F_a = 20 N[/tex] is the applied forward force

[tex]F_f=9.0 N[/tex] is the friction force

Combining the two equations,

[tex]F_a-F_f=ma[/tex]

And solving for the acceleration,

[tex]a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2[/tex]

Learn more about Newton's second law:

brainly.com/question/3820012

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