Question

A balloon weighing 'W' newton descends with an acceleration of 'a'. If weight 'w' is removed from the balloon, the balloon has upward acceleration of 'a'.
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w=2aW/(a+g)
g is acceleration due to gravity.

Answer 1

any luck y2o2?

Answer 2

Well, i'm trying..but i really like it :D

Answer 3

Don't worry , it'll be solved

Answer 4

ok! waiting!

Answer 5

are you sure there is nothing about the mass ?!

Answer 6

yeah sure, mass isnt mentioned anywhere.

Answer 7

mass

Answer 8

how about we eliminate mass after forming 2 equations?

Answer 9

nah, that gives W=2gw/a+g

Answer 10

The balloon has two forces acting on it. Gravity and the upward force form the gas in the balloon. Call the second \( F_b \). The first, \( F_g = -mg \) where \( m \) is the mass of the balloon. We're using the convention that upwards is positive and downward is negative. Hence \( F_b > 0 \) and \( F_g < 0 \).

The net force acting on the balloon is
\[ F_{net} = F_g(m_{initial}) + F_b \]
We're told that that initially the balloon is descending. Hence
\[ F_{net} = ma_{initial} \]
where \( a_{initial} < 0 \).

Answer 11

The weight \( W \) is the gravitational force, hence \( W = mg \). We're told that a weight of \( w \) is removed from the balloon. Hence the new weight of the balloon is
\[ -F_g = W' = W - w = mg - w \]
Hence the net force acting on the balloon now is
\[ F_{net}' = F_g' + F_b = -(mg - w) + F_b = w + (-mg + F_b) = w + F_net \ \ \ ---(1)\]
and we're told that now the balloon is now accelerating upward with the same magnitude of acceleration. Write \( -a = a_{initial} \). Therefore \( a \) is a positive quantity. Then we have
\[ F_{net} = -ma \ \ \ ---(2) \]
and
\[ F_{net}' = ma \ \ \ --- (3) \]

Now use the three equations--(1), (2) and (3)--to show the result you're looking for.

Answer 12

Equation one got chopped off. It is
\[ F_{net}' = w + F_{net} \ \ \ -- (1) \]

Answer 13

Actually, you'll need the more general form of equation (1)
\[ F_{net}' = w - mg + F_b \]

Anyway. Play around with those for a bit.

Answer 14

I solved it.............Finally

Answer 15

While going down, \[W - F_{air} = \frac{W}{g}a\]
After removal of mass w.
\[F_{air}-(W-w) = \frac{W-w}{g}a\]

Solve these now, eliminate \(F_{air}\).

Answer 16

After removal of weight* w.

Answer 17

Got it! Thanks y202, JamesJ, Ishan94 for you time and help. Much appreciated!!!

Answer 18

This was asked in the exam i am giving tomorrow, i hope they ask a similar one tomorrow :P