Function Posted June 28, 2014 Posted June 28, 2014 Hi everyone Here's one for my entrance exam for med school: "A man with a mass of m kg walks up three stairs with height h. The work delivered by the gravity is then?" I'd say -3mgh, but the correct answer is -mgh. Can someone please explain why? Thanks.
Sensei Posted June 28, 2014 Posted June 28, 2014 (edited) 3 stairs have total height h. Not single stair. If somebody would say "stair has 33.33 cm. Calculate energy needed to rise mass m, 3 stairs up." Then 3*33.33 cm = ~1 m And result would be m*g*1 m Edited June 28, 2014 by Sensei
Function Posted June 28, 2014 Author Posted June 28, 2014 Oh God no forgive me it was just a very tricky question. I'll cite in Dutch: "Jan heeft een massa m en loopt een trap op. Het hoogteverschil is h en de weg die Jan aflegt is 3h. De arbeid geleverd door de zwaartekracht is A. mgh B. 3mgh C. -mgh D. -3mgh" Translation: "Jan has a mass m and runs up some stairs. The height difference is h and the distance he travelled is 3h. The work delivered by the gravity is ..." Still think this is a doubtful question: what exactly is the height difference? Between 2 stairs or the total height difference he went through? Now it's easy to say that h is the total difference in height, which makes the work = -mgh But what about the actual displacement? Jan moves on a line that forms a specific angle with the ground: Right-angled triangle with height h, hypothenusa has length 2sqrt(2)h.. The man moves with an angle of (90+arccos(2/3 sqrt(2)))°, so I'd say that W = m*g*3h*cos(90 + ...)° Which obviously doesn't make sence. Help.
Sensei Posted June 28, 2014 Posted June 28, 2014 Distance 3*h is "c" in below Pythagoras triangle. And height is "a".
Function Posted June 28, 2014 Author Posted June 28, 2014 (edited) Well, yes, so sin(theta) = 1/3, and so the angle formed with the gravitational force-vector is 90° + arcsin(1/3).. but what now? since W = F * x * cos(theta) Okay hang on.. Be theta the angle formed with the horizontal line. tan(theta) = h/2h = 1/2 sin(theta) = h/3h = 1/3 cos(theta) = 2/3 Wanted angle is the angle formed between the displacement and gravitational force cos(wanted angle) = -sin(theta) = -1/3 Yay I got it W = F * x * (-1/3) = m*g*3*(-1/3) = -mgh Edited June 28, 2014 by Function
Sensei Posted June 28, 2014 Posted June 28, 2014 (edited) "The work delivered by the gravity is..." That doesn't care about distance in axis X something traveled, only by axis Y- height (distance from center of Earth). If c would be 1 km, answer still would be the same. Edited June 28, 2014 by Sensei
Function Posted June 28, 2014 Author Posted June 28, 2014 "The work delivered by the gravity is..." That doesn't care about distance in axis X something traveled, only by axis Y- height (distance from center of Earth). If c would be 1 km, answer still would be the same. Well yes but I think it's the same since your "c" is included in the cosine of the angle formed between the displacement and the force delivering the work, cancelling out "c" when including the cosine in the formula of work?
Endy0816 Posted June 29, 2014 Posted June 29, 2014 (edited) You are trying to answer in terms of total work done and the question is only asking about gravity. if the question were, "A man with a mass of m kg climbs up a ladder with height h. The work delivered by the gravity is then?" your answer would remain the same. Edited June 29, 2014 by Endy0816
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