Question

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.)

Answer 1

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

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

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.