Answer to Question #168072 in Mechanics | Relativity for buddy

Question

A motorcycle travelling 24 m/s comes to rest in 8.7 s. What is the force of friction of the road against the bike tires if the bike weighs 8,200 N?

Answer 1

$$F_{fr}\Delta t=m(v_f-v_i), ​$$ $$F_{fr}=\dfrac{m(v_f-v_i)}{\Delta t}=\dfrac{\dfrac{W}{g}(v_f-v_i)}{\Delta t},$$ $$F_{fr}=\dfrac{\dfrac{8200\ N}{9.8\ \dfrac{m}{s^2}}\cdot(0-24\ \dfrac{m}{s})}{8.7\ s}=-2308\ N.$$