How do I find energy lost? Something very simple is confusing me

AI Thread Summary
To find the kinetic energy lost, the correct method is to subtract the final energy from the initial energy, resulting in the equation Ek1 - Ek2. The initial kinetic energy was stated as -206 J, and the final as 10 J, leading to a calculation of -216 J. However, kinetic energy cannot be negative, indicating an error in the initial data. The discussion emphasizes that energy lost should be calculated as initial minus final, but the negative result suggests a misunderstanding in the values provided.
x86
Gold Member
Messages
256
Reaction score
18

Homework Statement


Find the kinetic energy lost


Homework Equations





The Attempt at a Solution


We solved it in class

Ek1 = -206 J
Ek2 = 10 J

Find the energy lost.

The answer is was 2.2 * 10^2 J lost

This is confusing for me, to find energy lost, we should subtract the initial energy from the final energy, or should we subtract the final from the initial? So is the answer

Ek1 - Ek2 = -206 J - 10 J = -216 J
or

Ek2 - Ek1 = 10 J - (-206 J) = 216 J

How do we do it?

I know for example if someone has say 10 marbles, and loses 3, we have seven left. So the amount lost should be

10 - 7 = 3

Or initial - final.

Or is it final - initial = 7 - 10 = -3 ?
 
Physics news on Phys.org
Your intuition with the marbles is correct. We have an initial situation which undergoes a process and then a final situation. So,

Energy lost = Initial - Final.

Although, Ill give you a standard disclaimer that 'lost' isn't the correct term to be using here.

I think your problem comes from your initial calculations. An object can't possesses a negative kinetic energy.

Ek = 1/2 m v^2

1/2 is +ve mass is +ve and velocity is either -ve or +ve, but (+ve)^2 = (-ve)^2 = +ve

Hence Ek is the product of three positive values and must always be positive (or 0).
 
2.2*10^2 J is 216 J written in scientific notation with 2 sig. figures.
 
If Ek1 is the initial energy and Ek2 is the final energy then the energy lost is

= EK1 - EK2
= (-206) - (+10)
= -216 Joules

It's negative => gain

Edit: Oops it's KE so I agree with what others have said, you can't have negative KE so something is wrong with the original data.
 
Last edited:
If KE1 is -206 J and KE2 is 10 then you have

KE1-KE2 = KE3

-206-10= -216

You are losing -216 J therefore:

-206 - (-216) = 10 J

But as pointed out by Gatsby88 you can't have negative kinetic energy.
 
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...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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...
Back
Top