Questions
‘n’ identical cubes each of mass ‘m’ and side ‘l’ are on the horizontal surface. Then the minimum amount of work done to arrange one on the other is
detailed solution
Correct option is C
work done w= mghWork done to put the second one on the top of the first mgl+l2-mgl2=mgl (center of mass of the cube has to be considered as the position of the center of mass) similarly for the third one to put on the second work done is mg2l+l2- mgl2=2mgl similarly for the next work done is 3mgl and so on upto n-1 terms so total work done is mgl(1+2+3+.......n-1) mgln(n-1)2=Talk to our academic expert!
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A bucket of mass ‘m’ tied to a light rope is lowered at a constant acceleration of ‘g/4’. If the bucket is lowered by a distance ‘d’, the work done by the rope will be (neglect the mass of the rope)
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