Question

Please offer a step by step answer of how you came to know each. It needs to be very detailed so I can come to understand the concepts better and reasoning as to why each step was done. Thanks!

-10 points My Notes A 150-kg crate rests in the bed of a truck that slows from 50 km/h to a stop in 12 s. The coefficient of static friction between the crate and the truck bed is 0.655 (a) Will the crate slide during the braking period? Explain your answer. (b) Give the range of stopping times for the truck that will prevent the crate from sliding.

Answer 1

m = 150 kg

u = 50 km/h = 13.8889 m/s

t = 12 s

v = 0 m/s

n = 0.655

acceleration (retardation) of the truck, a = (v -u)/t = (0 - 13.8889)/12 = -(125/108) m/s² = -1.1574 m/s²

a) forward force on crate = m|a| = 150 x 125/108 N = 173.61 N

max friction force possible = nmg = 0.655 x 150 x 9.8 = 962.85 N

now max friction force is more than forward force, crate will not slide

b) to prevent sliding,

nmg $\geq$ m|a|

=> ng $\geq$ u/t

=> t $\geq$ u/ng

=> t $\geq$ (250/18)/(0.655 x 9.8)

=> t $\geq$ 2.1637 s