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datadisha datadisha
wrote...
Posts: 19
Rep: 0 0
11 years ago
The coefficient of kinetic friction is 0.43
force being pulled is 150 N
Distance pulled is 11.5 m
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1 Reply

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wrote...
11 years ago
Hello


The horizontal pulling force is 150*cos30° = 129,9 N
The vertical pulling force is 150*sin30° = 75 N

The vertical component of the pulling force reduces the  weight of the crate:
The frictional force is 0,43*(mg - 150*sin30)
Ffric = 0,43*(37*9,81 - 150*sin30)
Ffric = 123,8 N

The work done by the friction force =
work = F*s = 123,8 N* 11,5 m = 1423,7 Nm

Regards
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