Physics Help Forum 2 simple (I hope) questions about energy transformation

 Energy and Work Energy and Work Physics Help Forum

 May 2nd 2016, 04:05 AM #1 Junior Member   Join Date: May 2016 Posts: 3 2 simple (I hope) questions about energy transformation I am trying to work out how to answer these 2 questions, and am a little lost. Any help will be most appreciated! ================================= Q1 ================================= During the rush hour, a moving staircase carries 120 commuters per minute up from an underground railway to street level. The railway is 10 m below street level and the average commuter weighs 500 N. The average power output of the motor which drives the staircase must be at least A 1 000 W B 5 000 W C 10 000 W D 50 000 W E 600 000 W ================================= Q2 ================================= An archer pulls the string back a distance of 0.50 m. The string exerts an average force of 300 N on the arrow as it is fired. The mass of the arrow is 0.15 kg. The maximum kinetic energy gained by the arrow is A 23 J B 150 J C 600 J D 2000 J E 6750 J.
 May 2nd 2016, 07:50 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Where are you having trouble? In both these questions you have to calculate the work done. Do you know how to do this? Then use work done = energy change. The answer to question 1 is C and to question 2 is B, what do you think?
May 2nd 2016, 08:45 AM   #3
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 Originally Posted by studiot Where are you having trouble? In both these questions you have to calculate the work done. Do you know how to do this? Then use work done = energy change. The answer to question 1 is C and to question 2 is B, what do you think?
Thanks a lot - I got the same answers as well.

What confused me was:

for Q1:
I used EP (i.e. potential energy) = mass x g x height

and then Average Power = EP/t

I assume this is correct?

I wasn't sure if EP/t is the AVERAGE power...

For Q2:
The height of the archer (or rather the height from which the arrow is released) is not mentioned, not allowing me to calculate the additional energy stemming from the potential energy (mass x g x height) that can be converted into kinetic once the arrow leaves the bow, adding to the total.

So the answer should really be:

Max Poss Kinteic Energy = (F x distance) + (mass x g x height)

and not just F x distance...

May 2nd 2016, 09:08 AM   #4
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 Originally Posted by jcocker for Q1: I used EP (i.e. potential energy) = mass x g x height and then Average Power = EP/t I assume this is correct? I wasn't sure if EP/t is the AVERAGE power...
Yes, correct. Two clarifications however - you are given the weight of a passenger in Newtons, not the mass, so the gain in potential energy is force x distance = W x h. Having calculated the change in PE per passenger in N-m, given 120 passengers per minute that's 1/2 passengers per second, so the power required in the equal to the change in PE per passenger multiplied by passengers per second, yielding a result of N-m/ss, which is the same as watts. See how the units work out?

 Originally Posted by jcocker For Q2: The height of the archer (or rather the height from which the arrow is released) is not mentioned, not allowing me to calculate the additional energy stemming from the potential energy (mass x g x height) that can be converted into kinetic once the arrow leaves the bow, adding to the total. So the answer should really be: Max Poss Kinteic Energy = (F x distance) + (mass x g x height) and not just F x distance...
True - you raise a valid point. Keep in mind that in general for introductory physics problems they really aren't trying to trick you, so if they fail to mention things like the archer's height don't worry about it too much. It's good that you recognize that this could affect the answer, but so can other things such as wind speed (launch an arrow upward in a hurricane and it will gain a huge amount of KE!) or whether the archer is stationary. So in general if factors such as these aren't mentioned, you can ignore them.

 May 2nd 2016, 09:08 AM #5 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 They are always trying to trick you And there is always an easier way the profs use to get the answer. Q1 Power is the work done in 1 second. 120 people lifted 10 m per minute = 120/60 = 2 persons lifted per second. Each person weighs (weight = force) 500N 2 persons weight 1000N work = force x distance 1000N x 10 m = 10,000 Joules in 1 second 1 Joule/second = 1 watt 10,000 J/s = 10,000 w Q 2 The mass is irrelevent. All the energy comes from the bow. work = force x distance = 300 x 0.5 = 150J All this appears as KE, the velocity achieved depends on the mass of the arrow, is there another question about this?
May 2nd 2016, 09:36 AM   #6
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 Originally Posted by studiot They are always trying to trick you And there is always an easier way the profs use to get the answer. Q1 Power is the work done in 1 second. 120 people lifted 10 m per minute = 120/60 = 2 persons lifted per second. Each person weighs (weight = force) 500N 2 persons weight 1000N work = force x distance 1000N x 10 m = 10,000 Joules in 1 second 1 Joule/second = 1 watt 10,000 J/s = 10,000 w Q 2 The mass is irrelevent. All the energy comes from the bow. work = force x distance = 300 x 0.5 = 150J All this appears as KE, the velocity achieved depends on the mass of the arrow, is there another question about this?

Q1: Just how I did it.
Q2: no, there is nothing else to the question, hence why I was confused ;-)

Many thanks - much appreciated!

 May 2nd 2016, 09:55 AM #7 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Maybe there was once another part to question 2, that was removed but the data was not. Or perhaps they were deliberately adding irrelevent information to see if students know what they are doing. Whatever keep up the good work.

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