You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

 μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

            x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

            a = 2 x / t²

            a = 2 0.5 /0.5²

            a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

  X axis

                fr - Wₓ = m a          (1)

  Y Axis  

               N- = 0

                N = W_{y}

We use trigonometry to find the components of the weight

            sin 45 = Wₓ / W

           cos 45 = W_{y} / W

           Wₓ = W sin 45

           W_{y} = W cos 45

The out of touch has the expression

             fr = μ N

             fr = μ W_{y}

We substitute in 1

             μ mg cos 45 - mg sin 45 = m a

             W_{y} = (a + g sin 45) / g cos 45

              μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

           a = -4 m / s²

            μ = 1- 4 / (9.8 cos 45)

            μ = 0.423

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