A 81-kg sprinter wishes to accelerate from rest to a speed of 11 m/s in a distance of 29 m . What coefficient of static friction is required between the sprinter's shoes and the track?

Answer 1

To accelerate the runner pushes through their shoes against the ground which in turn pushes back on the runner, according to Newton's 3rd Law. This force comes from friction between the runner's shoes and the ground.

This friction force is μN, where N is the normal force.

Since the gravitational force mg and N balance the full friction force is μmg.

By Newton's 2nd Law μmg = ma ⇒ μ = a/g.

The useful kinematic equation here is v_f² = v_i² + 2ad, where v_f is the final speed, v_i is the initial speed, and d is the distance traveled.

Since the runner starts from rest we have a = v_f²/(2d) ⇒ μ = v_f²/(2dg)

Answer 2

We first need to determine the acceleration required for the sprinter to reach 11 m/s in 29 m. Assuming constant acceleration, we will use the distance and velocity as functions of time:

s(t) = v₀*t+1/2at²

v(t) = a*t

We know that the sprinter is starting from rest (v₀=0), he wants to achieve 11 m/s, and accomplish that in 29 m.

29m = 1/2at²

11m/s=a*t

Substituting the velocity equation in for at in the distance equation, we get

29 m = 1/2(11 m/s)*t

Which gives a time of t=5.27 s

We can then calculate the acceleration easily.

Now that we know the acceleration required, we can use Newton's Second Law: F=ma.

So the total force required will be F=81kg*a, where a is the acceleration value we just found. This acceleration is due entirely to the frictional force between the runner's shoes and the track. The maximum frictional force is calculated by the product of the normal force F_n=mg and friction coefficient μ.

We can equate the accelerating force with the frictional force:

mgμ = ma

Consequently, solving for μ, we get

μ=a/g.

I leave it as an exercise to the reader to actually crunch the numbers.

Answer 3

μ = a / g

Let us find a = Acceleration , μ = Coefficient of Friction , g = acceleration due to gravity = 9.8m/s2

Using Vf2 - Vi2 = 2a*d

Vf = 11m/s

Vi = 0

d= 29 m

a= 11*11/(2*29 )

a= 121 /58 = 2.09 m/s2

μ = a / g = 2.09/9.8 = 0.21 ( No units) = coefficient of friction