Weather Balloon Question: Calculating Acceleration, Velocity, and Net Force

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The discussion revolves around calculating the physics of a weather balloon and its instruments. The balloon exerts an upward force of 98 N, resulting in an acceleration of 19.6 m/s² for the instruments. After 10 seconds of acceleration, the instruments are released, and their velocity at that moment is crucial for determining subsequent motion. The net force acting on the instruments after release is 49 N, accounting for gravity. Finally, the instruments will first experience downward velocity after reaching their peak height, which occurs when the upward velocity becomes zero.
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Any help would be greatly appreciated. I'm doing physics through correspondents, so I have nobody to explain this kind of thing to me.

Homework Statement


The instruments attached to a weather balloon have a mass of 5.0kg.
a) The balloon is released and exerts an upward force of 98 N on the instruments. What is the acceleration of the balloon and the instruments.
b) After the balloon has accelerated for 10 seconds, the instruments are released. What is the acceleration of the balloon and the instruments?
c) What net force acts on the instruments after their release?
d) When does the direction of their velocity first become downward?


Homework Equations


F=ma
W=mg
V=d/t
A= (vf-vi)/t or A=F/m
(I cannot think of any others needed...)

The Attempt at a Solution


a)
a=F/M
a= 98N/5.0kg
a= 19.6 m/s2

b)v=d/t
v=0/0
v=0m/s2
This part confused me. Do I get the current velocity of the balloon, and that is the velocity of the instruments at the moment of their release? Or do I have to figure out how fast the balloon got after 10s of acceleration, then use that number for the velocity?

c) W=mg
W=(50kg)(9.80m/s2)
W=490 N

d) I'm not sure how to even attempt this part. How do I go about doing this part? Or do I only answer in a sentence?
 
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ahrog said:
Any help would be greatly appreciated. I'm doing physics through correspondents, so I have nobody to explain this kind of thing to me.

Homework Statement


The instruments attached to a weather balloon have a mass of 5.0kg.
a) The balloon is released and exerts an upward force of 98 N on the instruments. What is the acceleration of the balloon and the instruments.
b) After the balloon has accelerated for 10 seconds, the instruments are released. What is the acceleration of the balloon and the instruments?
c) What net force acts on the instruments after their release?
d) When does the direction of their velocity first become downward?


Homework Equations


F=ma
W=mg
V=d/t
A= (vf-vi)/t or A=F/m
(I cannot think of any others needed...)

The Attempt at a Solution


a)
a=F/M
a= 98N/5.0kg
a= 19.6 m/s2
It's F_{net} = ma. What's F_{net}?
b)v=d/t
v=0/0
v=0m/s2
This part confused me. Do I get the current velocity of the balloon, and that is the velocity of the instruments at the moment of their release? Or do I have to figure out how fast the balloon got after 10s of acceleration, then use that number for the velocity?
You want the velocity of the balloon and the instruments, i believe, not the acceleration as your problem states. Velocity, time, and acceleration are related by a basic kinematic equation of motion.
c) W=mg
W=(50kg)(9.80m/s2)
W=490 N
of course you mean 5(9.8) = 49N.
d) I'm not sure how to even attempt this part. How do I go about doing this part? Or do I only answer in a sentence?
It's looking for the time it takes for the instruments to move up at release, and come to a stop, before now changing direction and falling down. It's the kinematic equations again.
 
a) I'm pretty sure I did this one right. I needed acceleration, so I changed around the F=ma question. Would Fnet be just the 98 N the balloon exerts, or do I have to subtract gravity and the weight of the instruments from that?

b) So find the velocity of the entire thing, balloon and instruments, even though they have just been released?

c)That's what I meant, I accidentally added a zero.

d) What equation would I use for that?
 
ahrog said:
a) I'm pretty sure I did this one right. I needed acceleration, so I changed around the F=ma question. Would Fnet be just the 98 N the balloon exerts, or do I have to subtract gravity and the weight of the instruments from that?
the lift force is 98 up; the gravity or weight force is 49n down. The net force is not 98.
b) So find the velocity of the entire thing, balloon and instruments, even though they have just been released?
Yes! Once you get the correct acceleration, how fast is the thing going after 10 seconds?
c)That's what I meant, I accidentally added a zero.
cool, I'll buy it.
d) What equation would I use for that?
the initial speed you get from part b. The final speed at the top of the instrumnet's ascent is zero (that'a when they start to fall down). What is the acceleration of the instuments now that the balloon is out of the picture (think about gravity and the acceleration of gravity).
 
a) Okay, for that I got 9.8 m/s2 of acceleration.

b) The velocity is 98 m/s.
c) The net force is 49 N.
d) And the time would be 10 seconds.

Do these sound right to you?
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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