AugustCrawl said:
The efficiency is defined in terms of total useful output/ total input.
Note that the units of work done and energy supplied are both Joules.
From dimensional analysis you can likely show an equivalence.
So if your motor has done 100J of useful work but has consumed 200J of Energy doing this then efficiency is 50% or 0.5
It looks like your experiment is about using the motor to raise a load over a height. In this case the useful output energy of the motor is the same gravitational potential energy which the load has gained in being raised over a height. You have probably used the same height for all your loads - do you have the value in metres?
To calculate the energy consumed by your motor first consider the power consumed by the motor - which is the product of the current and the voltage. So for each raise by multiplying the voltage by the current you have the power... multiply that power by the time taken to find the energy consumed over that time period... you need to average all your values first.
Once you have this average energy consumed over the same time period and height for each raise (10 raises with 10 different loads) you now have 10 values for the total energy consumed.
what you then do is consider for each of your 10 raises what the total energy gained by the load was... this is done by multiplying the gravitational acceleration constant 9.81 metres per second squared by the mass by the height the mass was raised over. GPE = mgh. You now have 10 values for the useful work done by your motor.
Now all that is left to find how efficiency changes is to divide the useful output by the total input for each value and then multiply by 100.
Voila, you now have calculated the efficiency as a function of changing the mass load.
You can fill in the blank right-most column with this information.
Do you have to write a report too?
THANK U SOO MUCH...Yes I HAVE to write a report :S
the height for all of them is 25cm
By the way... I was sent this by someone and I was wondering if you could also help me make sense of it, on the parts you may not have mentioned. I am really bad at these things. :/
''It has been a while since I've worked with this, but I would start at the end of the equation and work my way backwards - based on that sheet, efficiency is (work/energy). From that sheet, we know energy (in Joules) is (W * s). s is seconds and W is Watts.
You don't actually have it labelled, but the work (in Joules) is the weight(force - Newtons)*height(meters). You have the weight there in grams as well as an equation to transform it to Newtons. You don't have heights listed, so I assume the height was the same for each experiment?
Given that, you have to find the equation for Watt, and then just plug in the numbers - find the work based on the force. Find the watts based on the voltage and current. find the energy based on the watts and time, then find the efficiency based on the work and energy.''