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In Fig , a block of density rho = 800 k...

In Fig , a block of density `rho = 800 kg//m^(3)` floats face down in a fluid of density `rho_(f) = 1200 kg //m^(3)` . The block has height H = 6.0 m
(a) By what depth h is the block submerged ?

(b) If the block is held fully submerged and then released , what is the magnitude of its acceleration ?

Text Solution

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From Eq. , we know that the buoyant force has the magnitude `F_(b) = m_(f) g` , where `m_(f)` is the mass of the fluid displaced by the block.s submerged volume `V_(f)` , From Eq. (`rho = m//V)` , we know that the mass of the displaced fluid is `m_(f) = rho_(f) V_(f)` , We do not know `V_(f)` but if we symbolize the block.s face length as L and its width as W , then from Fig . we see that the submerged volume must be `V_(f) = L Wh` . If we now combine our three expressions , we find that the upward buoyant force has magnitude
`F_(b) = m_(f) g = rho_(f) V_(f) g = rho_(f) L Whg.`
Similarly , we can write the magnitude `F_(g)` of the gravitational force on the block , first in terms of the block.s mass m , then in terms of the block.s density `rho` and (full) volume V , and then in terms of the block.s dimensions L , W , and H (the full height) :
`F_(g) = mg = rho V g = rho L WHg `
The floating block is stationary . Thus , writing Newton.s second law for components along a vertical y axis with the positive direction upward `(F_("net" , y) = ma_(y))` , we have `F_(b)- F_(g) = m(0)` ,
or from Eqs. 14-20 and 14-21,
`rho_f L W hg - rhoL W Hg = 0 `, which gives us
`h = (rho)/(rho_(f)) H = (800 Kg//m^(3))/(1200 kg//m^(3)) (6.0 cm)`
`=4.0` cm
(b) The gravitational force on the block is the same but now , with the block fully submerged , the volume- of the displaced water is V = LWH . (The full height of the block is used . ) This means that the value of `F_(b)` is now larger , and the block will no longer be stationary but will accelerate upward . Now , Newton.s second law yields `F_(b) - F_(g) = ma` ,
or `rho_(f) L W Hg - rho L W Hg = rho L W Ha`, where we inserted `rhoLWH` for the mass m of the block .
Solving for a leads to
`a = ((rho_(f))/(rho) - 1) g = ((1200 kg//m^(3))/(800 kg//m^(3)) - 1) (9.8 m//s^(2))`
`= 4.9 m//s^(2)`
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