Question

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 FfFfF_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.)

Answer 1

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