The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1x=x1. Find _f, 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.)

Magnitude of the Frictional force = (mv₀²)/2x₁

Explanation:

For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.

According to Newton's first law of motion

Frictional force = (mass of the box) × (deceleration experienced by the box)

Let the mass of the box be m

Then,

Frictional force = ma

Then we can obtain the deceleration using the equations of motion

v² = u² + 2ax

u = Initial velocity = v₀ m/s

v = Final velocity = 0 m/s (since the box comes to rest at the end)

x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)

a = ?

v² = u² + 2ax

0 = (v₀)² + 2ax₁

2ax₁ = - v₀²

a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)

Magnitude of the Frictional force = ma = (mv₀²)/2x₁

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