A 1 Kg object is dropped from rest from a helicopter at an altitude of 2000 meters?

  • Thread starter Kaxa2000
  • Start date
A 1 Kg object is dropped from rest from a helicopter at an altitude
of 2000 meters. What is the velocity of the object just before it lands upon a thick bed of pillows? The top of the pillow stack is at zero altitude. If the restoring force
constant (spring constant) of the stack of pillows is 1 N/m, how far into the stack will the
object travel before it comes to a full stop? Neglect air resistance.
 

CompuChip

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Re: A 1 Kg object is dropped from rest from a helicopter at an altitude of 2000 meter

So you can start by calculating the final velocity. There are two ways to go here: one would be to use forces (Newton's laws, etc); and the other is using energies (conversion from potential to kinetic energy, etc).

Since the second part of the question includes a spring, I suggest you are supposed to use energies here?

So there are two relevant points: the point where the object is dropped and the point 2000 meters below it where it hits the pillows. As you know, the total energy
Etotal = Epotential + Ekinetic
is conserved (i.e. the same at these two points), where Epotential is the gravitational energy.
Can you give me the formulas for the two quantities on the right hand side and express them in given values and unknowns?
 
Re: A 1 Kg object is dropped from rest from a helicopter at an altitude of 2000 meter

Would it be

mgh + 1/2mv2

(1kg)(9.8m/s2)(2000m) + (1/2)(1kg)(v)2
 
Re: A 1 Kg object is dropped from rest from a helicopter at an altitude of 2000 meter

After you solve for v you get 200 m/s.

Do you set the KE = to the spring equation after that to find distance it goes into pillows?

1/2mv^2 = kX?
 
Re: A 1 Kg object is dropped from rest from a helicopter at an altitude of 2000 meter

almost--energy in a spring is of the same form=1/2kx^2 As written above you are equating energy and force.
 

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