How to Calculate the Speed of a Block Released from a Spring

  • Thread starter Thread starter robertmatthew
  • Start date Start date
  • Tags Tags
    Spring
Click For Summary
SUMMARY

The speed of a block with a mass of 1.20 kg released from a spring with a force constant of 2.20 N/m after being compressed by 0.420 meters can be calculated using the conservation of energy principle. The initial potential energy (PE) stored in the spring is given by Ue = 0.5 * k * d², which results in a value of 0.5 * 2.20 N/m * (0.420 m)². This potential energy converts entirely into kinetic energy (KE) as the block leaves the spring, represented by the equation E = 0.5 * m * v². The correct approach involves equating the initial potential energy to the final kinetic energy to find the speed of the block.

PREREQUISITES
  • Understanding of potential energy and kinetic energy concepts
  • Familiarity with the spring constant and Hooke's Law
  • Basic algebra for solving equations
  • Knowledge of energy conservation principles in physics
NEXT STEPS
  • Calculate the initial potential energy using the formula Ue = 0.5 * k * d²
  • Learn how to derive the speed of an object using kinetic energy equations
  • Explore the implications of frictionless surfaces in energy conservation problems
  • Study the relationship between spring compression and energy storage
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of spring-block systems in action.

robertmatthew
Messages
48
Reaction score
0

Homework Statement


A block of mass 1.20 kg is placed in contact with a spring with a force constant of k = 2.20 N/m. When released, the spring pushes the block and expands a distance of 0.420 meters. If the floor is frictionless, the speed of the block as it leaves the spring is ? m/s.


Homework Equations


Elastic energy is Ue = .5kd2
Energy in a block-spring system is E = .5kx2 + .5mv2


The Attempt at a Solution


Well the problem is I haven't been able to get anywhere, because in my obviously incorrect thinking, if the block has yet to be released, then all the energy is potential, so E = Ue, but if E = Ue, then .5kx2 = 0, which can't be right. So, in short, I really don't know what I'm doing here. Am I using the wrong equations or something?
 
Physics news on Phys.org
robertmatthew said:

Homework Statement


A block of mass 1.20 kg is placed in contact with a spring with a force constant of k = 2.20 N/m. When released, the spring pushes the block and expands a distance of 0.420 meters. If the floor is frictionless, the speed of the block as it leaves the spring is ? m/s.


Homework Equations


Elastic energy is Ue = .5kd2
Energy in a block-spring system is E = .5kx2 + .5mv2


The Attempt at a Solution


Well the problem is I haven't been able to get anywhere, because in my obviously incorrect thinking, if the block has yet to be released, then all the energy is potential, so E = Ue, but if E = Ue, then .5kx2 = 0, which can't be right. So, in short, I really don't know what I'm doing here. Am I using the wrong equations or something?
Wecome to PF!
When the block is not yet released, the spring is compressed by 0.420 m. So its initial PE is not 0.
 
I just solved it, but I think my problem was that I was calculating the initial potential energy while also calculating the final kinetic energy, but still using E = PE, so I ended up with zero on one side and .5mv^2 on the other. I'm not really sure why I did Kf and Ui at the same time, I guess I'm having an off day, haha. But thank you!
 

Similar threads

Replies
29
Views
3K
Replies
17
Views
2K
Replies
61
Views
4K
Replies
24
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
30
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
14
Views
4K