A box is on the back of a truck, the truck is moving at 91.5 km/h along a level road. If the coefficient of static friction between the box and the truck is 0.360, what is least distance that the truck can stop in without the box sliding?

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

step-by-step explanation:

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted