Is it possible for a balloon to rise up with an acceleration of 1 m/s²?

  • Thread starter Thread starter Ockonal
  • Start date Start date
  • Tags Tags
    Force Mass
AI Thread Summary
A balloon with a weight of 10.8 kN is falling with an acceleration of 1 m/s², and the goal is to determine how much mass must be removed for it to rise with the same acceleration. The calculations show that the balloon's mass is 1080 kg, derived from the weight formula P = m × g. When applying Newton's second law, the force needed to achieve the desired upward acceleration is calculated to be 9720 N. The discussion highlights that the balloon's mass remains constant regardless of its motion, and adjustments in mass are necessary to change its acceleration. Ultimately, the conclusion is that the balloon cannot rise with the same acceleration without altering its mass.
Ockonal
Messages
30
Reaction score
0

Homework Statement


A balloon which has weight 10.8 kN fall down with acceleration 1 m/s². What mass we have to pull of from balloon to make it(ballon) rise up with same acceleration.

Homework Equations


P = m×g
P = m(g ± a)
F =m×a

P = 10.8 kN = 10.8×10³ N
a = 1 m/s²
g = 10 m/s²
m₀ - ? (but answer is in Newtons, so we have to find weight `P`)

The Attempt at a Solution


The first we have to find balloon mass using weight:
P = m(g-a)
m = P / (g-a) => m = 1200 kg;

Write first Newtons law (Y axis show down):
Y: ma = mg - F
We have to find that `F`
F = -m(a-g) => F = -1200*(1-10) = 10800 N.
Fail! F = P, and it's impossible. In this case balloon should be static without any movement.

Anyway, my next idea is:

-ma = (m - m₀)×g - F

-ma *— because acceleration should change the direction to opposite (up)
We can't edit `F` but we can change mass.
With calculations I got:

m₀ = (F-ma) / g + m

BSOD


What's wrong?
 
Physics news on Phys.org
The balloon's mass is given by the formula P=m*g, the speed at which it's moving doesn't change its mass.

So, 10800=m*10
then m=1080kg

Using that value in your calculation gives us:

F= -m(a-g) => F= -1080*(1-10) = 9720N
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top