First slide

Potential energy

Question

Twenty identical cubical blocks each of mass 100gm and of side 10cm are lying on a horizontal surface. Work done in piling them one above the other is, (g = 10ms^–2)

Moderate

A

20 J
B

2 J
C

19 J
D

10 J

Solution

d₁ = Displacement of C.G. of block -1 = l, where l = length of side. d₂ = 2l, d₃ = 3l,....d₁₉ = 19l
$\large \therefore$ W = mgd₁ + mgd₂ + ......+mgd₁₉
= mgl (1+2+3+....19)
$\large =mgl.\frac {19(19+1)}{2}=190mgl$
= 190 x 0.1 x 10 x 0.1J= 19J

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