Asked by dayton
A 80kg crate Is pushed 35m across the floor with a force of 500n the coefficient of kinetic friction is 0.25 how much for is done on the crate by pushing it"
Answers
Answered by
Henry
Fc = m*g = 80kg * 9.8N/kg = 784 N. = Force of crate.
Fn = Fc = 784 N. = Normal.
Fk = u*Fn = 0.25 * 784 = 196 N. = Force
of kinetic friction.
Work = (Fap-Fk))* d = (500-196) * 35 =
10,640 J.
Fn = Fc = 784 N. = Normal.
Fk = u*Fn = 0.25 * 784 = 196 N. = Force
of kinetic friction.
Work = (Fap-Fk))* d = (500-196) * 35 =
10,640 J.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.