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

Answer:

10 kg

Explanation:

While balloon coming down :

Weight = mass × Gravity

= Mg

Given acceleration = g/2

If an object is moving in air , air resistance also take place

Since balloon is coming down,

Air resistance = Mg - M/2 -‐-‐- equation 1

While balloon is moving up :

Let x grams is removed

Weight = ( M -x )g

Given acceleration = g/3

Since balloon is moving up,

Air resistance = Mg - mg + M/3 - x/3 ---------- equation 2

Equating equation 1 and equation 2

Mg - M/2 = Mg - xg + M/3 - x/3

Mg + Mg - M/2 = - xg + M/3 - x/3

- M/2 + x/3 = - xg + M/3

x/3 + xg = M/3 + M/2

4xg/3 = 5M/6

8xg = 5M

x = 5 × 160/ 8 × 10

= 800/80

= 10 kg

Answer 2

Answer:

The mass to be removed from it for the balloon to be moving up with an acceleration of $\frac{g}{3}$ $ms$ is $10kg$.

Explanation:

While balloon coming down :

$Weight = mass$ × $gravity= Mg$

Given acceleration $= \frac{g}{2}$ $ms$

If an object is moving in air , air resistance also takes place.

Since balloon is coming downwards,

Air resistance $= Mg - \frac{M}{2}$ -- Let this be (i)

While balloon is moving up :

Let $x$ grams is removed

Weight $= ( M -x )g$

Given acceleration = $\frac{g}{3}$

Since balloon is moving up,

Air resistance $= Mg - mg + \frac{M}{3}- \frac{x}{3}$   --- Let this be (ii)

From (i) and (ii) :

$Mg - \frac{M}{2} = Mg - xg + \frac{M}{3} - \frac{x}{3} \\\\Mg + Mg - \frac{M}{2}= - xg + \frac{M}{3} - \frac{x}{3}$

$- \frac{M}{2} + \frac{x}{3} = - xg + \frac{M}{3} \\\frac{x}{3} + xg = \frac{M}{3} + \frac{M}{2}$

$\frac{4xg}{3} = \frac{5M}{6} \\8xg = 5M$

$x = 5 * \frac{160}{8} * 10\\= \frac{800}{80} \\= 10 kg$

The mass to be removed from it for the balloon to be moving up with an acceleration of $\frac{g}{3} ms = 10kg$.