Sum

An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6?

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#### Solution

Let U be the upward force of water acting on the plastic box.

Let m be the initial mass of the plastic box.

When the empty plastic box is accelerating upward,

\[U - mg = \frac{mg}{6}\]

\[ \Rightarrow U = \frac{7 mg}{6}\]

\[\Rightarrow m = \frac{6U}{7g} . . . . \left( i \right)\]

Let M be the final mass of the box after putting some sand in it.

\[Mg - U = \frac{Mg}{6}\]

\[ \Rightarrow Mg - \frac{Mg}{6} = U\]

\[ \Rightarrow M = \frac{6U}{5g} . . . . \left( ii \right)\]

Mass added

\[= \frac{6U}{5g} - \frac{6U}{7g}\]

\[= \frac{6U\left( 7 - 5 \right)}{35 g}\]

\[ = \frac{6U \cdot 2}{35 g}\]

From equation (i),

\[m = \frac{6U}{7g}\]

∴ Mass added

\[= \frac{2}{5}m\]

Concept: Newton’s Second Law of Motion

Is there an error in this question or solution?

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