Conservation of Energy in a Block-Spring System

AI Thread Summary
In a block-spring system with a constant spring constant and amplitude, the total mechanical energy is determined by the amplitude and spring constant, not the mass of the block. Given an initial energy of 24 J with an amplitude of 4.5 cm, replacing the block with one of double the mass does not change the energy, as energy is proportional to the square of the amplitude. The energy remains at 24 J because the parameters affecting energy (spring constant and amplitude) are unchanged. The misconception that doubling the mass would double the energy is clarified, emphasizing that energy is independent of mass in this context. Thus, the energy of the system remains 24 J.
PhysLiz
Messages
6
Reaction score
0

Homework Statement



A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 4.5 cm has an energy of 24 J. If the block is replaced by one whose mass is twice the mass of the original block and the amplitude of the motion is again 4.5 cm, what is the energy of the system?

Homework Equations



E= K+U = .5mv^2 + .5kx^2 = .5kA^2

The Attempt at a Solution



So if the spring constant doesn't change and the amplitude doesn't change, will the energy of the system stay constant even though the mass (m) is doubled? Initially I thought the energy of the system would be doubled, because I thought it made sense that it would take more energy to move a more massive object, but that was incorrect. Thanks in advance for any assistance
 
Physics news on Phys.org
Solved.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Distribution of Energy when work is done on a system of 2 masses connected by a spring
Replies
17
Views
1K
Why can we move the spring with constant speed when we apply a force?
Replies
29
Views
2K
A block dropping onto a spring system
2
Replies
58
Views
2K
Time period of SHM & Energy conservation in pulley-spring-block system
Replies
24
Views
3K
Does amplitude depend on mass in SHM?
Replies
1
Views
8K
Problem with when to use Force and Energy. What compresses spring/
Replies
3
Views
1K
Damped harmonic motion
Replies
3
Views
595
Trouble understanding the influence of a real spring in a system
Replies
3
Views
2K
Shm , calculation of amplitude of spring mass system
Replies
3
Views
2K
A block of mass m is dropped onto the top of a vertical spring....
Replies
8
Views
5K

Hot Threads

Recent Insights

Back
Top