Respuesta :
Answer:
 x = 25 / μ   [ ft]
Explanation:
To solve this exercise we can use Newton's second law.
Let's set a reference system where the x axis is parallel to the road
Y axis Â
    N_B + N_A - W_van - W_load = 0
    N_B + N_A = W_van + W_load
X axis
   fr = ma
   a = fr / m
the total mass is
    m = (W_van + W_load) / g
the friction force has the expression
   fr = μ N_{total}
   fr = μy (W_van + W_load)
we substitute
   a = μ (W_van + W_load)   [tex]\frac{g}{W_van + W_load}[/tex]
   a = μ g
taking the acceleration let's use the kinematic relations where the final velocity is zero
    v² = v₀² - 2 a x
    0 = v₀² -2a x
    x = [tex]\frac{v_o^2}{2a}[/tex]
    x = [tex]\frac{v_o^2}{2 \mu g}[/tex]
    x = [tex]\frac{40^2}{2 \ 32 \ \mu}[/tex]
    x = 25 / μ   [ ft]
