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Q & A: Calculating Efficiency

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Most recent answer: 05/24/2013
Q:
If a machine takes in 50 J (Joules) and puts out 45 J what is its effiency?
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- Julie (age 12)
tampa, florida, u.s.a
A:
Julie -

There are lots of different ways of describing efficiency, but I think that what you are looking for is percent efficiency. What this measures is how much of the energy that you put into a machine you can get back out. For example, if you put 100 Joules of energy into a machine, and got 50 Joules back out (and the other 50 Joules was wasted by the machine), you would have 50% efficiency.

You can calculate percent efficiency by taking how much energy you got out, dividing it by how much you started with, and multiplying by 100%. So, if you put 10 Joules in and got 4 Joules out, you would have:

% Efficiency = (4 J) / (10 J) * 100% = 40%

So, if you put in 50 Joules and got 45 Joules back, you would have:

% Efficiency = (45 J) / (50 J) * 100% = ?

-Tamara
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(published on 10/22/2007)

Follow-Up #1: calculating machine efficiency

Q:
well i calculated this and got 90% i did 45/50=0.9x100=90 sp is the effiency 90 percent
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- zakiyyah (age 14)
london/southall
A:
Yes. Tamara did a good thing in encouraging the previous reader to work through this following her example. We think enough time has passed for it to be harmless to say that you have done it correctly.

Mike W.
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(published on 04/13/2011)

Follow-Up #2: what does efficiency mean?

Q:
My physics teacher wants us to calculate the efficiency of out catapults. she gave us a formula, but it doesn't work. The mass of the tennis ball we are launching is 57g, and the horizontal distance it traveled is 16 meters. The length of the arm is .6 meters (2 feet). She said to multiply the mass and the distance, and then divide that by the length of the arm. I did that, but I got in the hundreds. What am I doing wrong?
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- Melanie Rayburn (age 17)
Howe, Texas, United States
A:

As we say in our question guidelines, we don't answer homework problems. For this one, we also don't understand what's being asked well enough to answer. That gives us an opportunity to give some ideas about how to approach problems like this in general, which may turn out to be more useful.

Before answering any question you have to figure out what it means.  Here we really don't know what is meant by a catapult's efficiency.  Usually "efficiency" means how much of something you get per how much of something you have to supply. 

What could you be trying to get from the catapult? Maybe it would be the energy of the ball. What would you have to supply? Maybe that's also energy, the energy you have to supply as work to load up the catapult and get it ready to launch. We don't know that's what's meant here, but it's just an example of what someone might mean. So in that case what you'd do is divide the energy the ball got by the energy you put in to get the efficiency.

Calculating the ball's energy isn't hard. Perhaps the teacher gave you some sort of guidelines toward how to calculate that. I don't know how your catapult is loaded, but with more description maybe you could figure out how much work you had to do to load it. Then, if "efficiency" happened to mean what we guessed, you could do the calculation.

That formula you mentioned, "multiply the mass and the distance, and then divide that by the length of the arm", gives some mass as the answer. Your answer probably has some units, maybe kilograms. For answers with units, the units are very important. An answer of "hundreds" of grams is the same as an answer of "a fraction of" a kilogram. Checking units is really important to see if an answer makes sense.  At any rate, that formula seems very odd for an "efficiency", but could  be what's being asked for, for reasons that are beyond us.

Often an efficiency is a pure number, without units. For example, the "efficiency" of an air conditioner is often taken to mean the ratio of the energy it pumps out of a room to the energy it draws from the electrical outlet. A good result for that might be, for example "4.3" That would mean that paying for 1 Joule of electrical energy is enough to pump 4.3 Joules of heat out of your house. (Unfortunately, those efficiencies are often given in some strange units in which electrical  energy is measured in different units than heat energy.)

Mike W. (posted without checking until Lee gets back)

p.s. If things didn't  get straightened out in class, maybe your teacher could follow up here.

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(published on 05/24/2013)

Follow-up on this answer.