A flatbed truck is loaded with a stack of sacks of cement whose combined mass is

Question

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is 1029.0 kg. The coefficient of static friction between the bed of the truck and the bottom sack in the stack is 0.372, and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to 54.9 mph in 24.2 s. The stack of sacks is 1 m from the end of the truck bed. Does the stack slide on the truck bed?
Yes ______ or No _______   
The coefficient of kinetic friction between the bottom sack and the truck bed is 0.257. What is the work done on the stack by the force of friction between the stack and the bed of the truck?
__________
thank you for your help

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is 1029.0 kg. The coefficient of static friction between the bed of the truck and the bottom sack in the stack is 0.372, and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to 54.9 mph in 24.2 s. The stack of sacks is 1 m from the end of the truck bed. Does the stack slide on the truck bed?
Yes ______ or No _______   
The coefficient of kinetic friction between the bottom sack and the truck bed is 0.257. What is the work done on the stack by the force of friction between the stack and the bed of the truck?
__________
thank you for your help
54.9 mph in 24.2 s. The stack of sacks is 1 m from the end of the truck bed. Does the stack slide on the truck bed?
Yes ______ or No _______   
The coefficient of kinetic friction between the bottom sack and the truck bed is 0.257. What is the work done on the stack by the force of friction between the stack and the bed of the truck?
__________
Yes ______ or No _______    Yes ______ or No _______    thank you for your help

Explanation / Answer

Given Mass, m = 1029.0 kg coefficient of static friction,s = 0.372 Initial velocity, vi = 0 m/s Final velocity,vf = 54.9 mph = 24.54 m/s Time taken , t = 24.2 s The acceleration of the truck, a = (vf-vi)/t = 24.54/24.2                                                                =1.01 m/s^2 Now we will find the value of k *g = 0.257 *9.8 m/s^2                                                                = 2.51 m/s^2 since a is less than the value of  k *g  the stacks does not slide   Coefficient of kinetic friction,k = 0.257 Work done on the stack by the force of friction is W = fr *s      = s * mg *s = 0.372 *1029 kg *9.8 m/s^2 *1 m      = 3751.32 J

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