A force f = ( 4.0 x i + 3.0 y j ) n acts on a particle which moves in x-direction from a origin to x = 5.0 m. find the work done on the object by the force.
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work done = integration ( force.displacement )
= Integration ( 4xi +3yj).(dxi)
= 4x^2 x varying from 0 to 5
= 4(5^2)
= 4*25 = 100 J
Hope you get it. :)
= Integration ( 4xi +3yj).(dxi)
= 4x^2 x varying from 0 to 5
= 4(5^2)
= 4*25 = 100 J
Hope you get it. :)
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Answer:
50j
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