Question

An empty plastic box of mass 5kg accelerates up at rate of guy 6 when placed in water how much sand must be filled so that it accelerates down with gby6

Answer 1

The weight of the sand that should be added will be 2 kg

Let the buoyant force acting on the box = F

since the box moves upward with acceleration = g/6

=> F - mg = mg/6

=> F = 7mg/6

Let M be the mass of sand added so that it moves downward with acceleration, g/6

=> (m+M)g - F = (m + M)g/6

=> F = 5(m+M)g/6

comparing both the eqn we get,

5(m+M)g/6 = 7mg/6

=> 5m + 5M = 7m

=> 5M = 2m = 2 x 5 = 10

=> M = 2 kg

Hence the weight of the sand will be 2 kg

Answer 2

Answer:

Let the air resistance force is Fa and Buoyant force is B.

Given that, Fa=av

where v= velocity

→FM=kv,

where k= proportionality constant

When the balloon is moving downward

B+kv=Mg ................i.

→m=$\frac{b+kv}{g}$

for the balloon to rise with a constant velocity  v(upward) ≤ the mass of m Here , B-(mg+kv)=0 B=mg+kv<br.rarr m=(B+kv)/g:.Amount of mass removed should be =M−m

=$\frac{b+kv}{g}$−$\frac{b-kv}{g}$

=$\frac{b+kv-b+kv}{g}$

=2$\frac{kv}{g}$=2$\frac{mg-b}{g}$

=2{m-$\frac{b}{g}$}

Explanation: