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The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)

Respuesta :

Answer:

[tex]Ff=m\times \dfrac{V_o^2}{2X_1}[/tex]

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

[tex]V^2=U^2-2aS[/tex]

[tex]0=V_o^2-2a X_1[/tex]

[tex]a=\dfrac{V_o^2}{2X_1}[/tex]

So the friction force on the box

Ff= m x a

[tex]Ff=m\times \dfrac{V_o^2}{2X_1}[/tex]

Where m is the mass of the box.

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