Solve Weather Balloon Acceleration & Velocity Questions

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SUMMARY

The discussion focuses on solving physics problems related to the acceleration and velocity of a weather balloon and its attached instruments. The correct acceleration of the balloon is determined to be 10 m/s², not 20 m/s² as initially calculated. After 10 seconds of acceleration, the instruments reach a velocity of 98 m/s at the moment of release. The net force acting on the instruments post-release is 49 N due to gravity, and the instruments begin to fall after approximately 10 seconds when their upward velocity is countered by gravitational force.

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  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of gravitational force calculation (W = mg)
  • Familiarity with kinematic equations (vf = vi + at)
  • Basic concepts of forces acting on objects in motion
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  • Study the application of Newton's Laws in real-world scenarios
  • Learn more about kinematic equations and their applications in physics
  • Explore gravitational force effects on free-falling objects
  • Review the principles of buoyancy and forces acting on balloons
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Since I have a textbook without that ans for a certain section, that's why I want to check did I do these questions in the right way. Please tell me the right answers and show me the steps if I got them wrong. Thank you :wink:
This is the question:

The instrument attached to a weather balloon have a mass of 5.0kg.

a. The balloon is released and exerts an upward force of 98N on the instruments,What is the acceleration of the balloon and instruments?

I got 20m/s^2

b. After the balloon has accelerated for 10 s, the instruments are released. What is the velocity of the instrument at the moment of their release?

I got 196 m/s

c. What net force acts on the instruments after their release?

I got 147N

d. When does the direction of their velocity first become downward?

When the instrument stopped going upward and start falling.

Once again, thank you so much for helping me :!)
 
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a. The 5kg mass has a weight of about 49N (W = mg), this is in a direction opposite to the lift of the balloon. Since they're in opposite directions, we can subtract them to find the resultant forces. 98 - 49 = 49 Newtons up.
F = ma, so the force we just worked out, 49 = 5x, x ~ 10. The acceleration is 10ms^-2.

b.) vf = vi + at, since the balloon started at rest, vi = 0. So vf = at, 9.8x10 = 98ms^-1.

c.) The net force is ONLY the gravity, as the balloon has stopped applying a force on the instruments. The force of gravity is about 49N (9.8 x 5). Remember, the force on an object doesn't matter what the speed is. An object could be going close to the speed of light, and have no force acting on it. All it means is that it's going constant velocity.

d.) The object when it's released still has velocity, which you worked out in question b.). So only when that velocity is gone from gravity, will it start to fall. It's quite hard to visualize. But just imagine throwing a ball from a car on the highway. It would bounce along in the direction that the car is moving.

So for this:
vf = vi + at
vf = 0, as we're finding the point when the instruments momentarily stop in the air.
vi = 100.
at = -9.8t (it's negative because we chose up as being positive, and down as being negative).

0 = 100 + -9.8t
9.8t = 100
100/9.8 = t = 10s.

About 10 seconds after the object has been dropped, it will start to fall down.
 
blue boy: note that you should show your work before you can receive help in the homework forums here. BlackWyvern: please note that full solutions should never be given to homework questions, and help should only be given after the OP has shown his/her work. It may help for both of you to check out the Forum Rules, available from the "rules" button on the top toolbar.
 

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