Work & Hooke's Law Homework: Find Speed of Mass at Equilibrium Position

In summary, A vertical spring with a spring constant of 1000N/m and a mass of 5.0kg placed on top is held compressed at 25cm from equilibrium position. When released, the mass reaches a speed of 1.64m/s at equilibrium position. However, the answer booklet states the speed should be 2.76m/s, indicating a potential error in the calculations.
  • #1
alingy1
325
0

Homework Statement


A vertical spring of negligible mass and spring constant k=1000N/m has a small object of mass M=5.0kg placed on its top. The spring is held compressed by distance of 25 cm from equilibrium position. The spring is released. Find speed of mass at equilibrium position.


Homework Equations



W spring=1/2kdeltax^2
W gravity=deltax*5*9.8
Total energy gain of mass=19J
Energy transferred into potential: 12.25J=mgh
Energy transferred into kinetic: 6.75J ---->v=1.64m/s


The Attempt at a Solution



The answer booklet says the answer is 2.76m/s!
Where did I go wrong?
 
Physics news on Phys.org
  • #2
I don't know. The 19J seems too low. I got about 32 J. You didn't subtract the potential energy change twice, did you?

Chet
 
  • #3
W spring=1/2kdeltax^2
When evaluating this, don't overlook that power of 2.
 

Related to Work & Hooke's Law Homework: Find Speed of Mass at Equilibrium Position

What is Hooke's Law?

Hooke's Law is a physical law that states the force required to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed.

How does Hooke's Law relate to work?

Hooke's Law is related to work because the work done on a spring is equal to the force applied multiplied by the distance the spring is stretched or compressed. This can be represented by the equation W = 1/2kx², where W is work, k is the spring constant, and x is the displacement of the spring.

What is the equilibrium position?

The equilibrium position is the point at which the forces acting on an object are balanced and the object is at rest. In the case of Hooke's Law, the equilibrium position is when the spring is neither stretched nor compressed.

How do you find the speed of a mass at the equilibrium position?

To find the speed of a mass at the equilibrium position in Hooke's Law, you can use the equation v = √(k/m)(A² - x²), where v is the speed, k is the spring constant, m is the mass, A is the amplitude, and x is the displacement from the equilibrium position.

What are the units for the spring constant?

The units for the spring constant, k, depend on the units used for force and displacement. In the SI system, the units for k are newtons per meter (N/m). In the English system, the units are pounds per inch (lb/in).

Similar threads

Hooke's Law vs. Conservation of Energy
    • Introductory Physics Homework Help
Replies
5
Views
2K
Hooke's Law Lab Spring Constant Calculation
    • Introductory Physics Homework Help
Replies
3
Views
1K
Derive the formula for the frequency of a spring
    • Introductory Physics Homework Help
Replies
3
Views
1K
I Is Hooke's law really not working, or is it me being dummy?
    • Classical Physics
Replies
13
Views
989
How to find the equilibrium position of three masses (two of them fixed)?
    • Introductory Physics Homework Help
Replies
4
Views
896
Why does Hooke's law not work here?
    • Introductory Physics Homework Help
Replies
3
Views
2K
No prefix: Understanding the Total Force on a Mass Attached by Two Springs
    • Introductory Physics Homework Help
Replies
4
Views
2K
Hooke's Law Second Order Differential Equation
    • Introductory Physics Homework Help
Replies
14
Views
5K
Solving Hooke's Law Troubleshooting Homework
    • Introductory Physics Homework Help
Replies
6
Views
978
Confused about Hooke's law when analyzed at different times
    • Introductory Physics Homework Help
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top