Answer:
the least distance the truck can stop is 91.554 m so the box do not slide
Step-by-step explanation:
following Newton's second law , if the box should not slide then the acceleration exerted to the box should not be greater than the static friction force , that is
F friction = m * a max
since F friction = μ*N
μ= coefficient of static force
N= force normal to the box
applying Newton's fist law for the vertical forces , we get
N= m*g
thus
F friction = m * a max
μ*m*g = m * a max
a max = μ*g
assuming that the truck has a constant deacceleration , and finishes with velocity v =0 , then
v final ² = v initial ² - 2*a*L
0 = v initial ² - 2*a*L
L = v initial ² / (2*a) = v initial ² / (2*μ*g)
replacing values
L = v initial ² / (2*μ*g) = (91.5 km/h* 1000 m/km * 1 h/3600 s)² /(2*0.360*9.8 m/s²) = 91.554 m
thus the least distance the truck can stop is 91.554 m so the box do not slide
