Question

7. A flatbed truck is loaded with a stack of cement whose combined masses is 1143.5 kg. The coefficient of static friction between the bad of truck and the bottom sack in the stack is 0.372, and the sack are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck acceierates unifommy from rest to 91.08 km/h in 22.9 s. The stack of sacks is 1 m from the end of the truck bed The coefficient of kinetic friction between the botiom sack and the ruck bed is 0.2 a) Does the sack slide on the truck bed? 7 Marks) What is the work done on the stack by the force of frietion between the stack bed 6 Marks) TOTAL: 13 MARKS b) the bed of the truck? END OF QUESTIONS HAKCIPTA TERPELIHARA USIM 4

Answer 1

Answer :

a) no the stack does not slide off,   b) work done by friction is -368,000 J

(a)

Velocity = 91.08km/h = 27.24 m/s

time = 22.9

By v = u + at
=>27.24 = 0 + a x 22.9
=>a = 1.189 m/s^2
By F(net) = F(applied) - F(friction)
=>m x a' = m x a - µs x m x g
=>a' = a - µs x g
=>a' = 1.189 - 0.257 x 9.8
=>a' = 1.189 - 3.65
=>a' = a -ve value
=>F(applied) < F(friction)
=>The bags will not move.

(b)

Work done on the stack by the force of friction is
W = fr *s
= μs * mg *s = 0.372 *1143 kg *9.8 m/s^2 *1 m
= 368,000 J