The discussion revolves around calculating the work done by an elevator motor in lifting a 500 kg elevator a height of 50 m. Participants are exploring the physics of work in the context of forces and displacement.
The discussion is active, with participants questioning the assumptions made about the angle in the work formula. Some guidance has been offered regarding the correct application of the work formula under the assumption of no acceleration.
There is a focus on understanding the relationship between force direction and displacement, with some participants expressing uncertainty about the implications of the angle in the work calculation.
Why is the angle between the force and the displacement 90o?dsptl said:How much work does an elevator motor do to lift a 500 kg elevator a height of 50 m?
attempt: W = F.S.Cos90 = 0 J
but i think its not right... Please help some1?
Well, in which direction is the force applied? And in which direction is the elevator moving?dsptl said:is tht suppose to be 0 then?
Assuming that the elevator does not accelerate, that would be correct.dsptl said:upwards...
so is tht going to something like W = mg x 50m x cos0