Calculating work done by friction

Click For Summary
To calculate the work done by friction when a spring is compressed and released, the distance used in the work formula W=Fs should only account for the distance traveled by the object after it surpasses the uncompressed spring position. The discussion highlights that if the surface under the spring is not frictionless, friction will affect the energy conversion. Initial velocity after release can be determined using conservation of energy principles, where the elastic potential energy (EPE) converts to kinetic energy (KE). The remaining energy after accounting for friction represents the work done by friction. Understanding these concepts is crucial for accurately solving related physics problems.
rleung3
Messages
18
Reaction score
0
Hi,

When you compress a spring and release it (allowing object to spring some distance), to compute the work done by friction, your s term in W=Fs would have to equal the distance that the spring is compressed + the additional distance traveled by the object once it leaves the spring, right?

That's what I alwasy thought, but in one of my problems, it uses an s value that equals ONLY the distance traveled by the object after it surpasses the point of the uncompressed spring.

Ryan
 
Physics news on Phys.org
Does it say the surface under the spring is frictionless?
 
No, it does not...
 
Do you know the initial velocity after the object is released? You could use conservation of energy to see how much EPE is converted to KE. Whats left over is the work taken out of the system by friction. Then you can add this to what you figure out later
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
View full post »

Similar threads

A block dropping onto a spring system
  • · Replies 58 ·
2
Replies
58
Views
3K
Distribution of Energy when work is done on a system of 2 masses connected by a spring
  • · Replies 17 ·
Replies
17
Views
2K
Finding the work done by a block
  • · Replies 4 ·
Replies
4
Views
3K
Calculating the spring force constant K
  • · Replies 14 ·
Replies
14
Views
2K
Why is the work done double its expected value? (conveyer belt)
  • · Replies 58 ·
2
Replies
58
Views
5K
What is the Work Done on the Person+Sled System in an Amusement Park Ride?
  • · Replies 1 ·
Replies
1
Views
1K
The Work of Friction: Explained in .32m
  • · Replies 8 ·
Replies
8
Views
2K
Solving for Work Done When Accelerating a Mass
  • · Replies 3 ·
Replies
3
Views
882
Work Done by Friction & Gravity on Incline: Explained
  • · Replies 7 ·
Replies
7
Views
8K
Finding the Coefficient of Friction in a Spring-Block Incline System
  • · Replies 2 ·
Replies
2
Views
2K