Question

A balloon with its contents weighing 160N is moving down with an acceleration of g/2 ms. The mass to be removed from it so that the balloon moves up with an acceleration of g/3 ms is

Answer 1

F = Upward Force due to balloon

Before removing mass
Mg - F = Ma
160 - F = Mg/2
160 - F = 160/2
F = 80 N

After removal of mass (m)
F - (M - m)g = (M - m)a
F - (M - m)g = (M - m)g/3
F - Mg + mg = Mg/3 - mg/3
4mg/3 = 4Mg/3 - F
m = 3/(4g) × [4Mg/3 - F]
m = 3/(4 × 10) × [(4 × 160 / 3) - 80]
m = 10 kg

Mass to be removed is 10 kg

Answer 2

Answer:

The mass to be removed from the balloon, so that it move upwards with an acceleration $\frac{g}{2}$ $m/s^{2}$ is $10 kg$.

Explanation:

From given data,

• Initially, the weight ($W = mg$)of balloon acting downwards (due to gravity), $W = 160N$.
• Let the mass of balloon be $m$ and acceleration be $a$.

• For a balloon, there will also be an upward force and let the upward force be $F$.

Thus, $ma =$ Net force.

Also, $g = 10 m/s^{2}$

Therefore, $m = \frac{W}{g}$

                      $= \frac{160}{10} \\= 16 kg$.

Case: 1 Balloon moving down

This implies, Net force $=$ downward force $-$ upward force.

Here, the acceleration, $a = \frac{g}{2} m/s^{2}$

On substituting,

$m(\frac{g}{2} ) = W - F$

$F = W - m(\frac{g}{2} )$

$F = 160 - 16(\frac{10}{2} )$

$F = 80 N$.

Case: 2 Balloon moving up

Let the mass to be removed, to move upward be $x$.

Thus, W $= (m - x)g$

             $= (16 - x)10$

             $= 160 - 10x$.

This implies, Net force = upward force - downward force.

Here, the acceleration, $a = \frac{g}{3} m/s^{2}$

On substituting,

          $(m-x)(\frac{g}{3} ) = F - W$

          $(m-x)(\frac{g}{3} ) = F- (m - x)g$

        $(16-x)(\frac{10}{3} ) = 80 - (16 - x)10$

        $(16 - x)(\frac{40}{3} ) = 80$

                         $x = 16 - 6$

                         $x = 10$ $kg$.

Thus, the mass to be removed from balloon, so that the balloon move upwards is $10kg$.