n ideanalytical cubes each of mass m and side l are on the horizontal surface. Then the minimum amount of work done to parents then one on the other is
1)nmgl
2)mgln*2/2
3)mgln(n-1)/2
4)mgln(n+1)/2
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Initially when all blocks are at horizontal suface then the potential energy of the system of blocksPEi=(nm)g⋅l2When all blocks placed one above other then the centre of mass of these blocks will be at nl2 so nowthe potential energy of the systemPEf=(nm)gnl2so work done will be W=PEf−PEiW=(nm)gnl2−(nm)g⋅l2W=mg2n(n−1)
so option (C) is correct.
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