Question

Fred (60 kg) is rollerblading at a velocity of 25 km/h [E] when he sees a broken glass bottle on the path ahead. He brakes and is slowed to a velocity of 8 km/h[E] in 4.2 s.What is the coefficient of friction between Fred's rollerblades and the ground? (Hint: Start by rearranging Ff=μFN into μ=Ff/Fn. Find Fn and Ff. Here F_NET = F_f , therefore F_NET = ma becomes F_f = ma)

Answer 1

Let's first find the acceleration of the Fred:

$$a=\dfrac{v-v_0}{t}=\dfrac{2.22\ \dfrac{m}{s}-6.94\ \dfrac{m}{s}}{4.2\ s}=-1.12\ \dfrac{m}{s^2}.$$

Let's apply the Newton's Second Law of Motion:

$$-F_{fr}=ma,$$ $$-\mu mg=ma,$$ $$\mu=-\dfrac{a}{g}=-\dfrac{-1.12\ \dfrac{m}{s^2}}{9.8\ \dfrac{m}{s^2}}=0.11.$$