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

`Ff=m* (V_o^2)/(2X_1)`

Step-by-step 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=(V_o^2)/(2X_1)`

So the friction force on the box

Ff= m x a

`Ff=m* (V_o^2)/(2X_1)`

Where m is the mass of the box.