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Speed dropping to zero when landing during a fall

  • Thread starter sgstudent
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  • #1
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Homework Statement


When a box falls in a vacuum, it accelerates at 10m/s^2 and if the mass is 10kg, then it has a net forfeit of 100N. As it falls for 10s, it has a final speed of 100m/s. At the moment of impact, there is a reaction force of 100N acted upon the box by the earth. But since the final speed where net force=0N is 100m/s, then why does the speed drop to 0? Since there is no other force to bring the speed down (decelerate) to 0m/s, so I'm unsure.

Homework Equations



Net F=ma

The Attempt at a Solution


I'm totally not sure what could bring the speed to zero since even if a body has 0 net force, it can have a high speed. And the only way to bring it to zero is if I have a external force to decelerate it to 0 speed first. Thanks for the help!
 

Answers and Replies

  • #2
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Hi sgstudent!

The Attempt at a Solution


I'm totally not sure what could bring the speed to zero since even if a body has 0 net force, it can have a high speed. And the only way to bring it to zero is if I have a external force to decelerate it to 0 speed first. Thanks for the help!
Not sure what you mean. There is an external force by the earth on the block as the impact takes place.
 
  • #3
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Hi sgstudent!



Not sure what you mean. There is an external force by the earth on the block as the impact takes place.
But isn't the external force only the same magnitude as the weight of the object? So in that case won't the object continue at the same speed with no acceleration? Because in a car, if the car is moving at constant speed, that means that the car has no net force. So to stop it, the car must have another additional force to push it in the other direction deceleration it. But in the case of falling, it does not have another additional force to bring its speed down to zero. Thanks for the help Infinitum!
 
  • #4
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But isn't the external force only the same magnitude as the weight of the object?
The magnitude of force exerted by the earth due to impact is given as,

[tex]F = \frac{\Delta mv}{\Delta t} = \frac{mv-0}{\Delta t}[/tex]

And that is not always equal to mg.

I also noticed something in the question

At the moment of impact, there is a reaction force of 100N acted upon the box by the earth. But since the final speed where net force=0N is 100m/s
This is not true always.
 
  • #5
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The magnitude of force exerted by the earth due to impact is given as,

[tex]F = \frac{\Delta mv}{\Delta t} = \frac{mv-0}{\Delta t}[/tex]

And that is not always equal to mg.

I also noticed something in the question



This is not true always.
Oh, so the normal upwards force acting on the object is initially greater than the object's weight to drop its speed to zero? Does it mean that that force is very large as it is almost instantaneous
 
  • #6
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Oh, so the normal upwards force acting on the object is initially greater than the object's weight to drop its speed to zero? Does it mean that that force is very large as it is almost instantaneous
Yes. This is correct.

If the earth were totally rigid, the box would have to deform, with the front end of the box coming to a stop first, followed by cross sections higher up. The details of the deformation would depend on whether the box behaves elastically or not. If the deformation were totally elastic, the box would bounce up after the compression wave passed from the bottom to the top, and then got reflected as a tensile wave back down to the bottom again.
 
Last edited:
  • #7
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Oh, so the normal upwards force acting on the object is initially greater than the object's weight to drop its speed to zero? Does it mean that that force is very large as it is almost instantaneous
Yep! Chet nails it perfectly :smile:
 
  • #8
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Yep! Chet nails it perfectly :smile:
Oh, so we have to incorporate momentum to help us to answer the question? Because sometimes we are given similar questions like to describe a sky diver's falling. When the parachute opens there is a net upwards force in order to slow him down. Until the speed reaches zero which is at the ground. But in this case the ΔT is a lot larger and the upwards force is not constant? But they are also using the same principals? Lastly, what will the ΔT be in thus case? Cos have not learnt about momentum yet. Thanks for the help!
 
  • #9
881
40
Oh, so we have to incorporate momentum to help us to answer the question? Because sometimes we are given similar questions like to describe a sky diver's falling. When the parachute opens there is a net upwards force in order to slow him down. Until the speed reaches zero which is at the ground. But in this case the ΔT is a lot larger and the upwards force is not constant? But they are also using the same principals? Lastly, what will the ΔT be in thus case? Cos have not learnt about momentum yet. Thanks for the help!
That equation above is called the impulse equation, and is generally applicable to impulsive forces(collision). For the parachute problem, the force might be varying as frictional force usually varies proportional to the speed of the object, and so on. You cannot apply the impulse-force equation as this is not a collision situation. You should learn about it soon :smile:

If you are still curious, read up in your text about this and also look at Khanacademy or MIT OCW for a good idea.
 
  • #10
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That equation above is called the impulse equation, and is generally applicable to impulsive forces(collision). For the parachute problem, the force might be varying as frictional force usually varies proportional to the speed of the object, and so on. You cannot apply the impulse-force equation as this is not a collision situation. You should learn about it soon :smile:

If you are still curious, read up in your text about this and also look at Khanacademy or MIT OCW for a good idea.
Oh how interesting, looks like I only scratched the surface of forces since there's so much relation between momentum and force. Thanks for the help Infinitum!
 

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