A uniform rectangular marble slab is 3.4m and 2.0m wide.it has a mass of 180kg.its originally lying on surface flat ground with It's 3.4m*2.0m surface facing up.How much work is needed to stand it on It's short end?(hint,think about its centre of gravity)
5 answers
m g h = m g (3.4/2)
Answer is right.
Work done = Change in potential Energy. Initially it was flat. so h = 0. Now when it is standing, the centre of gravity is at height of 3.4/2
Work done = Change in potential Energy. Initially it was flat. so h = 0. Now when it is standing, the centre of gravity is at height of 3.4/2
Please give me full answer
Work done = ∆P.E
W=(mhg)f - (mgh)i
(mgh)i =0, since it was lying down
And, h = ½(h of the marble)
h= ½(3.4)=1.7m
W= 180 kg × 9.8 m/s² × 1.7m
W= 2,998.8 J
W = 3.0 kJ
W=(mhg)f - (mgh)i
(mgh)i =0, since it was lying down
And, h = ½(h of the marble)
h= ½(3.4)=1.7m
W= 180 kg × 9.8 m/s² × 1.7m
W= 2,998.8 J
W = 3.0 kJ
ty sa solution and formula
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