How can the force of impact on a falling spring mass system be calculated?

In summary: When m and k1 hit the ground, the spring will rebound with the same kinetic energy as before, and the total force on the system is the sum of the two original forces: mg Newtons.
  • #1
chandran
139
1
i have a spring with stiffness k. The spring is resting on the ground. Now a mass is kept over the spring. Now the spring deforms by an amount x=mg/k
where g is the gravity's acceleration.The force acting on the spring is mg Newtons and the reaction force from the ground is also equal to mg Newtons
since this is a statics problem.


Now i take the same spring and the mass over it and drop it over another spring of stiffness k1 fixed to the ground. The spring mass has an initial velocity of v.Now how can i calculate the force exerted on this system by the ground due to the impact.

After calculating this suppose i find that the force of impact is equal to x times mg can is replace the mass by a mass equal to xm and do a static analysis to find deflection.
 
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  • #2
are you asking a question?
 
  • #3
Now i take the same spring and the mass over it and drop it over another spring of stiffness k1 fixed to the ground. The spring mass has an initial velocity of v.Now how can i calculate the force exerted on this system by the ground due to the impact.
You'll have to clarify the interaction between the springs.

When the mass m, falls with the spring, there is no 'weight' pushing down on the spring, since the mass and spring are accelerating with gravity. So while falling the spring k is not deflected, i.e. there is not stored energy.

When the mass, m, and spring of constant k, hit the spring of constant k1, the mass and spring have an initial velocity and kinetic eneryg. Assuming an elastic response, both k and k1 will deflect, each absorbing some of the kinetic energy of m and k (assuming the spring has some mass). Then its a matter conservation of energy.
 

1. What is a falling spring mass system?

A falling spring mass system is a physical system that consists of a mass attached to a spring, which is in turn connected to a fixed point. The system experiences gravitational force due to the mass and a restoring force due to the spring.

2. What factors affect the motion of a falling spring mass system?

The motion of a falling spring mass system is affected by the mass of the object, the stiffness of the spring, and the gravitational force acting on the object. The initial conditions, such as the initial position and velocity, also play a role in determining the motion.

3. How is the motion of a falling spring mass system described?

The motion of a falling spring mass system can be described using Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In the case of a falling spring mass system, the acceleration is determined by the gravitational force and the restoring force of the spring.

4. What is the equation of motion for a falling spring mass system?

The equation of motion for a falling spring mass system is given by F = mg - kx, where F is the net force, m is the mass of the object, g is the acceleration due to gravity, k is the spring constant, and x is the displacement from the equilibrium position.

5. How does the energy of a falling spring mass system change over time?

In a frictionless system, the energy of a falling spring mass system remains constant. However, in real-world systems, energy is lost due to friction and air resistance. As the object falls, its potential energy decreases while its kinetic energy increases. When the object reaches the equilibrium position, all of its potential energy is converted into kinetic energy. As the object continues to move, its kinetic energy decreases and is converted back into potential energy as it reaches the opposite end of the motion.

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