Question

A general Physics Question regarding Buoyant Force.

If a block of 5kg mass is immersed in water attached to a spring of K constant of 500Kg in position 1, and the Buoyant force pushes the block to a certain high H which will be position 2, so the spring is now stretched, how could the change in H be found?

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Answer 1

The free body diagram for the given block can be written as shown below.

Buyoant force Block Force due to spring 'Spring Weight of the block

The equation for the above diagram can be written as

BWB + spring

where buoyant force fB, weight of the block WB and fspring are calculated using

spring =-k×H

fluid density of the fluid Vfluid - Volume of the fluid displaced by the block mB-mass of the block g - acceleration due to gravity at that place k - spring constant H-the upward displacement of the block

but, density of fluid times volume of fluid gives the mass of the block m_B. The numerical value of spring constant is not clear, thus I could not calculate the actual value of the upward displacement of the block, H.

$m_{B}\times - \ g =m_{B} \times g - k \times H$

the buoyant force is acting in the upward direction, but the acceleration due to gravity is acting in the downward direction. Thus - g is considered while taking the value of buoyant force.

still If we substitute the values of m_B as 5 kg, k as 500 N / m, g - 9.8 m s^-2,

thus

2x5x9.8 = 0.196 m ●H= 500

Please let me know the correct value of acceleration due to gravity at the place and the value of the spring constant with proper units, to calculate the proper value of H.