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ans is m'=0.035 kg
a cubical block of wood of edge 3cm floats in water.The lower surface of cube just touches the free end of a vertical spring fixed at the bottom of pot.Find max weight that can be put on the block without wetting it.Density of wood=800 kg/m3 and spring constant=50N/m
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UP 0 DOWN 0 0 2
2 Answers
106Draw the FBD in that condition:
kx upwards, mg of block + m'g of weight downwards and B buoyant force upwards.
kx = 50*(3*10-2) = 1.5 N upwards
mg of block = VÏg = (3*10-2)3*800*10 = 0.216N downward
m'g = m' is to be calculated
B = VÏwg = (3*10-2)3*1000*10 = 0.27N upward
Considering equilibrium,
1.5-0.216-10m'+0.27=0
or m' = 0.1554 kg
Calculation mistakes possible