Spring Force and kinetic energy

Click For Summary
SUMMARY

The discussion focuses on calculating the kinetic energy of a 0.0109 kg mass object bouncing on a spring with a force constant of 1.34 N/m, oscillating 3.49 cm above and below its equilibrium position. The correct formula for kinetic energy (KE) is KE = 0.5 * m * v², while the potential energy (PE) in the spring is given by PE = 0.5 * k * x². Participants clarify that the mass should not be squared in the kinetic energy formula, and emphasize the importance of using the correct coefficients in the equations.

PREREQUISITES
  • Understanding of basic physics concepts such as kinetic energy and potential energy.
  • Familiarity with Hooke's Law and spring constants.
  • Ability to manipulate and solve algebraic equations.
  • Knowledge of units of measurement, particularly in the context of mass and distance.
NEXT STEPS
  • Review the derivation of the kinetic energy formula KE = 0.5 * m * v².
  • Study the principles of Hooke's Law and its applications in spring mechanics.
  • Explore the conservation of energy in oscillatory systems.
  • Practice solving problems involving spring potential energy and kinetic energy calculations.
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding oscillatory motion and energy conservation in spring systems.

CMATT
Messages
31
Reaction score
1
A novelty clock has a 0.0109 kg mass object bouncing on a spring that has a force constant of 1.34 N/m.

How many joules of kinetic energy does the object have at its maximum velocity if the object bounces 3.49 cm above and below its equilibrium position?
 
Physics news on Phys.org
CMATT said:
A novelty clock has a 0.0109 kg mass object bouncing on a spring that has a force constant of 1.34 N/m.

How many joules of kinetic energy does the object have at its maximum velocity if the object bounces 3.49 cm above and below its equilibrium position?
This sounds like a homework question.
 
haruspex said:
This sounds like a homework question.
It is, I think I put it in the homework forum, I didn't know that was a thing until after I posted this
 
CMATT said:
A novelty clock has a 0.0109 kg mass object bouncing on a spring that has a force constant of 1.34 N/m.

How many joules of kinetic energy does the object have at its maximum velocity if the object bounces 3.49 cm above and below its equilibrium position?

Ive done (.05)(force constant)(mass)^2 and I still haven't gotten the correct answer.
Anyone know how to get it?! Please show me :(
 
CMATT said:
Ive done (.05)(force constant)(mass)^2 and I still haven't gotten the correct answer.
Anyone know how to get it?! Please show me :(
Mass2? Where did that come from? This is not gravitational attraction.
 
haruspex said:
Mass2? Where did that come from? This is not gravitational attraction.

(.05)(k)(m)^2 isn't an equation? first I was doing (.05)(k)(m), got it wrong, and then my friend said to square the mass, however it is still wrong
 
CMATT said:
(.05)(k)(m)^2 isn't an equation? first I was doing (.05)(k)(m), got it wrong, and then my friend said to square the mass, however it is still wrong
Oh dear.
Look up your notes, or some websites. What is the law for the force in a stretched (or compressed) spring? What is the equation for the potential energy?
 
haruspex said:
Oh dear.
Look up your notes, or some websites. What is the law for the force in a stretched (or compressed) spring? What is the equation for the potential energy?

I know KE = (.05)(m)(v)^2
KE = (.05)(m)(v)^2 +mgh

Spring PE = (.05)(k)(x)^2
x: the distance?
k: force constant
 
CMATT said:
I know KE = (.05)(m)(v)^2
KE = (.05)(m)(v)^2 +mgh

Spring PE = (.05)(k)(x)^2
x: the distance?
k: force constant
Yes, much better.
Suppose that the spring extension is A at the equilibrium position, and the amplitude of oscillation is A (=3.49cm, but keep all the working symbolic for now).
What will be conserved as the mass rises from its lowest point to the equilibrium position? Can you write that as an equation?
 
  • #10
Those are probably just typos but you need to fix (.05) to (0.5) in both equations.
 
  • #11
gmax137 said:
Those are probably just typos but you need to fix (.05) to (0.5) in both equations.
Well spotted.
 

Similar threads

Replies
29
Views
3K
Replies
17
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
20
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
2K