Question

The steepest street in the world is Baldwin Street in Dunedin, NZ. It is inclined at an angle of 380 , with the horizontal. A child slides down the street with a constant velocity on a sled with high friction runners. What is the coefficient of friction between the sled runners and the street?

Answer 1

Newton's second law allows to find the result for the friction coefficient of the street is:

• The friction coeficinwete is: μ = 0.78

Newton's second law establishes a relationship between the net force, the mass and the acceleration of the body.

           ∑ F = m a

Where bold indicates vectors, m is to mass and acceleration.

In the attached we see a free body diagram, it is a diagram of the forces without the details of the body, the x-axis is parallel to plane also shown with the positive in the direction of movement, going down the plane and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight.

        Sin θ = $\frac{W_x}{W}$  

        cos θ = $\frac{W_y}{W}$ / W

        Wₓ = W sin θ

         $W_y$  = W cos θ

We write Newton's second law for each axis.

y-axis

         N- $W_y$  = 0

         N = mg cos θ

x-axis

        Wₓ - fr = ma

Since they indicate that the body goes down at a constant speed, the acceleration is zero.

         W sin θ = fr

The friction force is the macroscopic representation of the interactions between the two surfaces and the formula.

         fr = μ N

we substitute.

          fr = μ mg cos θ

         mg sin θ = μ cos θ  

        μ = tan θ

Let's calculate.

         μ = tan 38.0

         μ = 0.78

In conclusion using Newton's second law we can find the results for the friction coefficient of the street is:

• The frivtion coefficient is:  μ = 0.78

Learn more about the coefficient of friction here:  brainly.com/question/11808898