Fast quick Energy Conservation Question

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Energy conservation applies specifically to conservative forces, where mechanical energy (kinetic plus potential) is conserved. In contrast, when non-conservative forces, such as friction, are present, mechanical energy is not conserved as it transforms into other energy forms like heat and sound. Despite this, total energy remains conserved in all scenarios. Therefore, while mechanical energy can be lost due to non-conservative forces, the overall energy in a closed system is always preserved. Understanding these distinctions is crucial for analyzing energy transformations in physical systems.
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For energy conservation does it apply to both conservative and non conservative forces?
 
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godkills said:
For energy conservation does it apply to both conservative and non conservative forces?

Mechanical energy (kinetic + potential) is conserved only if conservative forces act. If non-conservative forces act, then the sum of potential + kinetic energy may not be conserved. However, total energy is always conserved. It's just that when non-conservative forces act, the energy is converted into forms other than kinetic or potential.
 
Take friction for example. The kinetic energy is lost in the form of sound and heat. Total energy remains conserved but mechanical energy is lost.
 

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...
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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...
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