N identical cubes each of mass 'm' and side 'l' are on the horizontal surface.Then the minimum amount of work done to arrange them one on the other
Answers
Answer:
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Answer:
The minimum amount of work done to arrange N identical cubes one on the other is Nmg(N-1)(L/2).
Explanation:
Given,
Mass of N identical cubes = m
Length of a side of N identical cubes = l
Work done is the change in potential energy.
W = ΔPE = Pf - Pi
where,
Pf = Final potential energy
Pi = Initial potential energy
Pi = N×m×g×L/2
The total mass, when the N cubes are placed on one another is Nm and the total height is nL. So,
Pf = N×m×g×NL/2
Therefore,
Work Done = Pf - Pi
W = N×m×g×NL/2 - N×m×g×L/2
W = N×m×g×(N-1)×L/2
Hence, the minimum amount of work done to arrange N identical cubes one on the other is Nmg(N-1)(L/2).
To learn more about the work done, click on the link below:
https://brainly.in/question/93876
To learn more about the potential energy, click on the link below:
https://brainly.in/question/2769084
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