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While you push a box you begin to decrease the force you are exerting on the box. When will the box reach equilibrium?

A. When the force you are exerting on the box is equal to 0.

B. When Fyou on box = -Fbox on you.

C. When the force of the box becomes Fyou on box < Fbox on you

D. The box will never reach equilibrium.

While you push a box you begin to decrease the force you are exerting on the box When will the box reach equilibrium A When the force you are exerting on the bo class=

Respuesta :

The application of a force on the box overcomes the friction and causes the box to accelerate

While the force is decreasing, the box will stabilize;

  • B. When [tex]\underline{F_{you \, on \, box} = -F_{box \, on \, you}}[/tex]

Reason:

Newton's third Law of Motion states that action and reaction are equal in magnitude and opposite in direction

When the box is pushed with a force, [tex]F_{you \, on \, box}[/tex], the reaction of the box of mass, m,  is as follows;

  • [tex]F_{you \, on \, box}[/tex] = [tex]F_{friction}[/tex] + m × acceleration

Where;

[tex]F_{friction}[/tex] = [tex]-F_{box}[/tex] for example

The negative sign is due to the direction of the friction force, or [tex]F_{box}[/tex] is opposite to the direction of the applied force

When the force is decreased to Fâ‚‚ = [tex]F_{you \, on \, box}[/tex], such that we have;

[tex]F_{you \, on \, box}[/tex] = [tex]-F_{box \, on \, you}[/tex] , which is the condition for equilibrium, and the net

force is zero and according to Newton's first law of motion,  the box moves

at constant speed no acceleration

Therefore the box will reach equilibrium when;

  • [tex]\mathbf{\underline{F_{you \, on \, box} = -F_{box \, on \, you}}}[/tex]

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