A 8.0\,\text {kg}8.0kg8, point, 0, start text, k, g, end text box is released from rest at a height y_0 =0.25\,\text my 0 =0.25my, start subscript, 0, end subscript, equals, 0, point, 25, start text, m, end text on a frictionless ramp. The box slides from the ramp onto a rough horizontal surface. The box slides 2.0\,\text m2.0m2, point, 0, start text, m, end text horizontally until it stops.

μ = 0.125

Explanation:

To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.

Let's start working on the ramp

starting point. Highest point of the ramp

         Em₀ = U = m h y

final point. Lower part of the ramp, before entering the rough surface

        = K = ½ m v²

as they indicate that there is no friction on the ramp

          Em₀ = Em_{f}

          m g y = ½ m v²

          v =

we calculate

          v = √(2 9.8 0.25)

           v = 2.21 m / s

in the rough part we use the relationship between work and kinetic energy

          W = ΔK = K_{f} -K₀

as it stops the final kinetic energy is zero

          W = -K₀

The work is done by the friction force, which opposes the movement

          W = - fr x

friction force has the expression

          fr = μ N

let's write Newton's second law for the vertical axis

         N-W = 0

         N = W = m g

we substitute

            -μ m g x = - ½ m v²

           μ =

Let's calculate

           μ =

           μ = 0.125

hujjujddf

neither accurate nor precise.