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Why is is that only the change in vertical distance of the object is used to calculate work done
[From: ] [author: ] [Date: 12-03-24] [Hit: ]
.. [because cos90° = 0]=> your statement that 80 J of work is done in moving the object horizontally is incorrect.-You have the mathwrong and the formulas mixed up. The total work done is going to be mad.work is mad,......
=> work done = 20 x 4 x cos90° = 0 ... [because cos90° = 0]
=> your statement that 80 J of work is done in moving the object horizontally is incorrect.
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You have the math wrong and the formulas mixed up. The total work done is going to be mad. work is mad, easy to remember. It will depend upon how the mass is accelerated over time and the path taken in moving the mass. When the move is completed, the object will have gained a potential energy of mgh. If the mass is moved in two steps, say up and then to the right, the problem is much simplified. The mass has to be accelerated the right, as well as up.
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Think of the horizontal motion as being done on a frictionless surface. You can move the object as far as you wish without any force beyond that needed to get it moving (Newton's 1st Law) that force to get it moving is the only work in the horizontal direction. Thus, you do no work in the horizontal direction. To do work in the vertical direction you must lift the object
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it depends on the distance....the object is considered lifted on both cases,not lifted upwards and dragged to the right...
upwards+to the right=60+80=140
to the right +upwards=80+60=140
upwards=60...only3m
to the right=80...more,4m
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