Work done by the spring on an object

AI Thread Summary
An ideal spring is extended by a mass of 2.0 kg, initially stretching 6.0 cm and then an additional 10 cm due to an external force. The work done by the spring during this extension can be calculated using the potential energy formula, Fsp = 1/2 kx^2. Understanding the relationship between work and the change in potential energy is crucial for solving the problem. The discussion emphasizes the importance of correctly applying formulas and definitions related to work and energy in this context. Ultimately, a clear grasp of these concepts is necessary to determine the work done by the spring accurately.
DrHughes
Messages
5
Reaction score
0
20) An ideal spring is hung vertically from the ceiling. When a mass 2.0 kg hangs at rest from it, the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10cm. While the spring is being extended by the force, the work done by the spring is:

a) – 4.2 J
b) – 3.6 J
c) – 3.4 x 10-5 J
d) 3.6 J
e) 4.2 J

I know formulas like Fsp = 1/2 kx^2 etc .etc.
I still don't know how to put them to use in this context
 
Physics news on Phys.org
The formula you wrote is for the potential energy of the spring - not the force. This is not a good start. You'll have to make a better start before people can begin to help you.
 
If you want you can use energy to solve this question. Do you know how work and the change of potential energy are related?
 
DrHughes said:
I know formulas like Fsp = 1/2 kx^2 etc .etc.
I still don't know how to put them to use in this context
There's nothing wrong with using this formula for spring PE, if you know what it means. (Check your text!)

You can also attack the problem directly by considering the definition of work.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...

Similar threads

Replies
58
Views
2K
Replies
3
Views
1K
Replies
2
Views
4K
Replies
13
Views
5K
Replies
2
Views
1K
Replies
4
Views
1K
Replies
5
Views
2K
Replies
7
Views
1K
Back
Top