Calculating Work and Energy in a Spring-Block System

Click For Summary
SUMMARY

The discussion focuses on calculating work and energy in a spring-block system involving a block of mass 4.00 kg attached to a spring with a spring constant of 500 N/m. The block moves on a horizontal table with friction, initially compressed by 3.00 cm and moving at 0.600 m/s. Key calculations include determining the net work done on the block, the work done by the spring, and the work done by friction, ultimately leading to the calculation of the coefficient of kinetic friction.

PREREQUISITES
  • Understanding of Hooke's Law and spring potential energy (PE = 1/2 kx²)
  • Basic principles of work and energy in physics
  • Knowledge of kinetic friction and its coefficient
  • Ability to apply conservation of energy principles
NEXT STEPS
  • Calculate the net work done on the block using the work-energy theorem
  • Determine the work done by the spring during the decompression phase
  • Analyze the work done by friction to find the coefficient of kinetic friction
  • Explore the implications of non-conservative forces in mechanical systems
USEFUL FOR

Students studying physics, particularly those focused on mechanics, as well as educators and anyone interested in understanding energy transformations in spring-block systems.

gpber94
Messages
1
Reaction score
0

Homework Statement


A block with mass m = 4.00 kg is attached to a spring with spring constant k = 500 N/m, and it slides on a horizontal table that is not frictionless. At a time when the spring is compressed by 3.00 cm, the block is observed to be moving toward the right (decompressing the spring) at 0.600 m/s. When the block reaches the point where the spring is completely relaxed, the block is measured to be moving at 0.550 m/s. (a) What is the net work done on the block during the time interval? (b) What is the work done by the spring? (c) What is the work done by friction? (d) What is the coefficient of kinetic friction between the block and the table?


Homework Equations





The Attempt at a Solution


I honestly am lost, i know how to do a, but not anything after
 
Physics news on Phys.org
The net work done by the spring is given by the change in the spring's potential energy between two positions (the compressed one and the relaxed one), spring potential is
PE = \frac{1}{2}kx^2
Where x is the displacement from the relaxed position.
Since you know the total work and the part of the work done by the spring parts (c) and (d) follow :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
17
Views
2K
Replies
29
Views
3K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
24
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 27 ·
Replies
27
Views
10K