Maximum Velocity of a Falling Ball on a Spring

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SUMMARY

The maximum velocity of a 4 kg sphere falling from a height of 70 cm onto a spring with a constant of 200 N/m is calculated to be 4 m/s. The total energy conservation principle is applied, where the gravitational potential energy (Ep) is converted into elastic potential energy (Eel) and kinetic energy (Ek). The calculations involve using the equations Ep = mgh, Ek = mv²/2, and Eel = kx²/2, leading to the conclusion that the sphere's velocity upon contact with the spring is 4 m/s, confirming the energy conservation approach.

PREREQUISITES
  • Understanding of gravitational potential energy (Ep = mgh)
  • Knowledge of kinetic energy (Ek = mv²/2)
  • Familiarity with elastic potential energy (Eel = kx²/2)
  • Concept of energy conservation in mechanical systems
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Explore the dynamics of spring-mass systems
  • Learn about the relationship between force, mass, and acceleration (F = ma)
  • Investigate real-world applications of potential and kinetic energy in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of energy transformations in systems involving springs and falling objects.

Patolord
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Homework Statement


A sphere of 4kg falls from a height of 70cm and lands on a spring of constant 2 *10² N/m.
Whats the max velocity the sphere reachs?
g=10m/s²

Homework Equations


Ep=m*G*h
Ek=mv²/2
Eel=k*x²/2
Vmax=sqrt(2g*h) (internet source idk if this is right)

The Attempt at a Solution


I tried this Ep = Eel + epx
Ep=4*0,7*10
28=kx²/2 + mg*x
x=0,367 m
them i did 0,7 + 0,367 for the height
and it gave me Sqrt(2g*h)=v
V=4,62 m/s
not the answer, i couldn't get my head around the energies. I also tried Ep=Eel but no sucess
 
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Patolord said:
Ep=m*G*h
Ek=mv²/2
Eel=k*x²/2
These are OK.

Vmax=sqrt(2g*h) (internet source idk if this is right)
This only applies to a body falling freely from a height h. Don't need it.


The Attempt at a Solution


I tried this Ep = Eel + epx
Ep=4*0,7*10
28=kx²/2 + mg*x
x=0,367 m
them i did 0,7 + 0,367 for the height
and it gave me Sqrt(2g*h)=v
V=4,62 m/s
not the answer, i couldn't get my head around the energies. I also tried Ep=Eel but no sucess
The total energy is (using your notation): Ep + Ek + Eel
That total energy is conserved.

Hint: At what point does the mass stop accelerating? (After that point, it starts slowing down.)
 
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Sorry, power went out on my house.
So it stops acelerating when it touches the spring.
But i still need the energy of the spring right because it touches it?
If i try ep+ek + Eel i get 28 -2v² = 200x²/2
now to find the x
i can use the weight of the ball F=m*a = k *x , i get 40=200*x => x=0,2
them Eel = k*x²/2 = 4 !
Eel+ Ep = Ec Ec = 32
32= 4 * v² /2
16 = v²
v=4 m/s Yeaah lol solved while writing this thank you veryy much ! (or maybe i did something wrong and got to the right result ?)
 

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