A block of density pb=9.50×10^2 kg/m^3 floats face down in a fluid of density pf

Question

A block of density pb=9.50×10^2 kg/m^3 floats face down in a fluid of density pf=1.30×10^3 kg/m^3. The block has height H=8.00 cm.
a.) By what depth have is the block submerged?
b.) If the block is held fully submerged and then released, what is the magnitude of its acceleration?

4. (25 pts) A lock of density p 9.50 x 102 kg/m3 floats face down in a fluid of density Pi = 1.30 × 103 kg/ma. The block has height H-8.00 cm. (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?

Explanation / Answer

(a) Volume of the block (V)= AH
where A is area and H is its height
Volume of block submerged (v)= Ah
where h is the depth in fluid.
Now the downward force = weight of the block = (pb*V)g
Upward force = buoynat force = pf*g*v
Since the block is floating therfore both force must be same
(pb*V)g = pf*g*v
9.5*102*A*H = 1.3*103*A*h
h = (9.5*100 / 1.3*1000)*8 = 5.85 cm
Hence the depth which will be in fluid = 5.85 cm
(b) If block is fully submerged then the force on the block in upper direction
= pf*(AH)g
And the force in the downward direction = pb*g*(AH)
Hence the net force = pf*(AH)g - pb*g*(AH) = (AH)g*(pf - pb)
We know the net force = ma
where m is mass of block and a is acceleration
m = pb*AH
ma = pf*(AH)g - pb*g*(AH) = (AH)g*(pf - pb)
pb*AH* a = pf*(AH)g - pb*g*(AH) = (AH)g*(pf - pb)
a = [g*(pf - pb)] /pb
a = 3.614 m/s2

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