A block is submerged in vessel filled with water by a spring attached to the bottom of the vessel. In equilibrium, the spring is compressed. The vessel now moves downwards with acceleration$a(

Question

A block is submerged in vessel filled with water by a spring attached to the bottom of the vessel. In equilibrium, the spring is compressed. The vessel now moves downwards with acceleration$a( < g)$. The spring length [image:10_image67.jpeg]

Dr. Abubacker · Accepted Answer

(c) Let $k$ be the spring constant of spring and it gets compressed by length $x$ in equilibrium position. Let $m$ be the mass of the block and $F$ be the upward thrust of water on block. When the block is at rest,
$w = kx + F$ or $w - F = kx$
[image:10_image102.emf]
When the vessel moves downwards with acceleration $a\ ( < g)$ the effective downward acceleration$= g - a$. Now upthrust is reduced say it becomes $F''$
Where $F^{''} = \frac{F}{g}(g - a)$
In figure, then
$w - kx^{''} - F^{''} = ma$
or $w - kx^{''} - \left( \frac{g - a}{g} \right)F = \frac{wa}{g}$
or $(w - F) - kx^{''} + \frac{a}{g}F = \frac{wa}{g}$
or $kx - kx^{''} + \frac{a}{g}F = \frac{wa}{g}$
or $x^{''} = x + (F - w)\frac{a}{gk}\ $
hence, the spring length will increase

Q. A block is submerged in vessel filled with water by a spring attached to the bottom of the vessel. In equilibrium, the spring is compressed. The vessel now moves downwards with acceleration$a( < g)$. The spring length

Explanation

Reviewed by Dr. Abubacker

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