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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?

Respuesta :

KingHarzh
Fnet=m*a
a= Fnet/m

a=(20N[E]+9.0N[E])/ 12.0kg

a=2.4167m/s^2

a ≈ 2.4m/s^2

Answer:

The acceleration of the box its 0,91 m/s²

Explanation:

According to Newton first law, the force applyed to some object its equal to the mass of the object plus the acceleration of the same object.

   F=m.a ; clearing the acceleration of the equation ⇒ a=F/m

  Now in the problem its apeared the friction force and according to de phsysics definition its a force thats its against of movement.

  So the force its going to be equal to F= 20N - 9N.

 Replacing in the equation a=(20N - 9N)/12kg.

 Solving ⇒ a ≅0, 91 m/s²

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