Asked by Julius
A crate of mass 50kg is pulled up a rough inclined plane ,inclined at an angle of 60 degree to the horizontal.determine the total work done in moving the crate to the top of the inclined 200 meter hight.(show workin).
Answers
Answered by
Henry
Wc = m*g = 50kg * 9.8N/kg = 490 N = Wt.
of crate.
Fc = 490N @ 60o = Force of crate.
Fp = 490*sin60 = 424.4 N. = Force
parallel to incline.
Fv = 490*cos60 = 245 N. = Force perpendicular to the incline.
Ff = u*Fv = u*245 = Force of friction.
Fap-Fp-Ff = m*a
Fap-424.4-u*245 = m*0 = 0
Fap = 424.4 + 245u. = Force applied.
L = 200m/sin60 = 231 m. = Length of
incline.
Work = Fap * L=(424.4+245u) * 231.
of crate.
Fc = 490N @ 60o = Force of crate.
Fp = 490*sin60 = 424.4 N. = Force
parallel to incline.
Fv = 490*cos60 = 245 N. = Force perpendicular to the incline.
Ff = u*Fv = u*245 = Force of friction.
Fap-Fp-Ff = m*a
Fap-424.4-u*245 = m*0 = 0
Fap = 424.4 + 245u. = Force applied.
L = 200m/sin60 = 231 m. = Length of
incline.
Work = Fap * L=(424.4+245u) * 231.
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.