Scarpa Discount, Work, Energy and Power. PDFJun 3, 2010 — kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. . Substitution gives APE=W= (60. kg)(9.81 m/s²)(4.0 m).1 pageQuestions & answersBrainly.comQuestion17. A 60.-kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. Approximately how much total work is done against gravity by the student during the climb?Answer · 0 votesThe work done against gravity by the student is 2,352 J.Work doneThe work done on an object is the product of applied force and the displacement of the object.The given parameters:Mass of the student, m = 60 kgVertical distance, h = 4 mTime of climb, t = 8 sThe work done against gravity by the student is calculated as follows;W = P.E = mghW = 60 x 9.8 x 4W = 2,352 JLearn more about work done here: https://brainly.com/question/8119756MoreAssignment ExpertQuestionA 60 kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. Approximately how much total work is done against gravity by the student during the climb?Answer · 0 votesThe answer to the question is available in the PDF file https://www.assignmentexpert.com/https://www.assignmentexpert.com/homework-answers/physics-answer-21031.pdfMoreGauthmathQuestion5. A 60.-kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. Approximately how much total work is done against gravity by the student during the climb? A 2.4 * 103J B 2.9 * 102J C 2.4 * 102J D 3.0 * 101JAnswer · 198 votesAnswer: .MoreStudy.comQuestionA 60 kg student climbs a flight of stairs in 0.25 min. Each step is 18 cm and there are 20 steps.Answer · 0 votesThe student's average power is the rate of change of gravitational potential energy Ug=mgh.Here is what we know:• The student weighs m=60 kg.• The gravitational acceleration is g=9.8 m/s2.• The vertical displacement is h=20 steps×18 cmstep×m100 cm=3.6 m.• The duration is t=0.25 min×60 smin=15 s.The student's power P in Watts is determinable as follows:P=Ugt=mght=60 kg(9.8 m/s2)(3.6 m)15 s=141.12 WThe student's output power is about 140 W.Using the conversion 1HP=745.7 W i.e. k=HP745.7 W, calculate the power in horsepower as follows:P=141.12 W×HP745.7 W≈0.189 HPThe student's power is about 0.190 hp.MoreStudy.comQuestionA 60-kg student climbs a ladder a vertical distance of 4.0 m in 8.0 s. How much total work is.Answer · 0 votesGiven that a 60 Kg student climbs a ladder a vertical distance of 4 m. When he climbs the ladder he is doing work against the force of gravity (Fg=mg) that is trying to pull him down. Work done is by definition the force times the displacement ( s ) of the point of action of the force, if the force and displacement are parallel. W=Fg s=mgs.Since the displacement of the student is s=4 m , the total work he does aginst gravity is :W=mgs=60×9.8×4orW=2352 JMoreNumeradeQuestionA 6o.-kilogram student climbs a ladder a vertical distance of 4.0 meters in 8 . o seconds. Approximately how much total work is done against gravity by the student during the climb? (A) 2.4 ×10^3 J (B) 2.9 ×10^2 J (C) 2.4 ×10^2 J (D) 3.0 ×10^1 JAnswer · 4 votesOkay. In this question, we are asked to 60 kg students aligned by the vertical distance of four m in 10 seconds approximately how much the network is done against gravity by the student during the pipeline, As must we have given a 60 kg and At distance we have given is four m and time we have given is equal to here's again, so well done. We have to find out is to find out the debate which is equal to the force. We use the formula MG. And then we have given a 60 And the G is 10. Then we get it as 600 Newton. It means the forces 600 Newton. So therefore, well done is able to force into or displacement. So we're done. We get is equal to 600. Multiply by Before. So well done, we get is 2.4 Multiplied by 10 days to go. threeMoreCheggQuestionA 60kg student climbs a ladder a vertical distance of 4m in 8s. Approximately how much total work is done against gravity by the student during the climb?Answer · 0 votesWork d.MoreQuoraQuestionA boy of mass 60 kg climbs vertically up a 5m ladder in 3 seconds. What is the power of the boy?Answer · 1 voteQUESTION:A boy of mass 60 kg walks vertically up a 5 m ladder in 3 seconds. What is the power of the boy?ANSWER:It is impossible to determine the mean power (rate of energy expenditure) of the boy from the given information; the best we can do is to place a lower bound on that value by calculating the power expenditure involved in raising the boy's mass of 60 kg by 5 m in 3 seconds.Let E be the energy required raise the boy's mass by 5 m. This is the energy required to increase the potential energy of the boy from his original position to a position 5 m higher and is therefore equal to the difference between the boy's potential energy at the higher position and his potential energy at the lower position, which in turn is equal to the product of the (gravitational force F acting on the boy) and 5 m. F = mg, where m is the boy's mass and g is the gravitational acceleration, which is approximately 9.8 m/s².So F = mg = (60 kg)×(9.8 m/s²)= (60×9.8) kg×m/s² = 588 N.So E = F×(5 …MoreTopprQuestionA man of mass 60 kg climbs up a 20 m long staircase to the top of a building 10 m high. What is the work done by him? (Take g = 10 m s^-2 ).Answer · 88 votesHere, m = 60 kg , height h = 10 m Work done W = change in potential energy = mgh - 0 = mgh = 60 × 10 × 10 = 6000 J = 6 kJMoreTopprQuestionA man weighing 60 kg climbs up a staircase carrying a load of 20 kg on his head.The staircase has 20 steps each of ht 0.2 m. If he takes 10 s to climb. Find his power.Answer · 104 votesTotal work done by man in covering 20 steps = mgh = 80×10×0.2×20 = 3200JSo, Power(P) = Work doneTime taken = 320010 = 320 WattMore Feedback Scarpa Gea 24.5, WS - Energy 1H - work KEY. PDFA 60.-kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. 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