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juana slides a crate along the floor of the moving van the coefficient of kinetic friction between the crate and van floor is 0...Asked by Mark
juana slides a crate along the floor of the moving van the coefficient of kinetic friction between the crate and van floor is 0.120 the crate has a mass of 56.8 kg and juana pushes with horizontal force of 124 N. if 74.4 J of work are done on crate and the delta x is 1.3m. what is the final speed if the crate starts at rest?
Answers
Answered by
drwls
Here's a hint:
(Work done by Juana) - (Work done against friction)
= (Increase in kinetic energy)
Use the final kinetic energy to get the speed.
The friction force is
Ff = M*g*(0.12) = 66.8 N
The work done against friction is
Ff*1.3 = 209.6 J
The work done pulling is 124N*1.3m = 161.2 J
The difference becomes kinetic energy
(Work done by Juana) - (Work done against friction)
= (Increase in kinetic energy)
Use the final kinetic energy to get the speed.
The friction force is
Ff = M*g*(0.12) = 66.8 N
The work done against friction is
Ff*1.3 = 209.6 J
The work done pulling is 124N*1.3m = 161.2 J
The difference becomes kinetic energy
Answered by
maryam
1.3
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