Work done on box from gravitational force

  • Thread starter frozen7
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  • #1
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When lift up the box to 2m forward. Find out the work done by the gravitational force.

I find it by W = mg (x)
But my lecturer encourage me to do it in this way: W = 2mgsin(a), and told me that my method is correct in trigonometric but not really correct in physics.

I would like to know whether my method is really not suitable or it is correct actually?
 

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  • #2
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frozen7 said:
When lift up the box to 2m forward. Find out the work done by the gravitational force.

I find it by W = mg (x)
But my lecturer encourage me to do it in this way: W = 2mgsin(a), and told me that my method is correct in trigonometric but not really correct in physics.

I would like to know whether my method is really not suitable or it is correct actually?
Work done will remain equal to mgh where h is the height raised.However this expression can be written in other different ways , for example you can write h in terms of the distance transversed upwards and the angle of the inclined plane with the horizontal , which is what you did.Both methods are correct.Actually it depends on the question itself which method to use.In the questions either the height raised will be given or the angle & the length transversed upwards will be given .

BJ
 
  • #3
mukundpa
Homework Helper
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Here the box is displaced agains the gravity, hence the work done by the gravitational force should be negative.
 

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