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potential energy mystery in published paper

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I'm trying to determine how much potential energy I need to generate via rubber bands to propel a mechanism a certain distance off of the ground. I've started by using mgh which seems straightforward. I.e. if I know my mechanism weighs "m" kg and I want it to jump "h" meters I just need to use m x 9.8 x h to determine how many Joules I'll have to generate with my bands.

And assuming I know the spring constant for my particular rubber band and how far I can stretch it, I can use 1/2 kx^2 to determine how many rubber bands I'll need to generate the energy.

At this point you're probably wondering "why the heck did this guy post?"

The problem is that I've been using this published paper as inspiration. One of the very first passages seems to contradict my assumptions about how to calculate the potential energy needed for my mechanism.

Here's the passage:

"...the jumping module needs to have an energy-storing capacity of more than 46.7 J/kg (1.4 J in 30.0g module) to allow the whole system to jump to height of 2m, assuming that the mass of the whole system is about 60.0g"

My question is this:

Assuming a mass of 0.060kg and a desired height of 2m, shouldn't the energy-storing capacity of the aforementioned module be mgh or (0.06kg) (9.8m/s^2) (2m) = 1.176 J?

The calculation I've made above doesn't line up with what the paper states. I've tried to figure out how the author is arriving at 46.7J/kg and none of the formulas I'm familiar with gives me any clue.

I'd really like to understand what I'm doing wrong - mostly to learn. If anyone here can explain how and why the author arrives at his number it would be greatly appreciated!

asked May 10, 2017 in Physics Problems by gronk33 (120 points)

1 Answer

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Best answer

Your calculation is correct.

The reason for the discrepancy is probably related to the design of the jumping mechanism, and to the fact that stored energy cannot be converted to kinetic energy with 100% efficiency. The joints of the rhombus-shaped legs move inwards horizontally as well as outwards vertically, but only the vertical motion is useful - see eg Fig 8 or 11. So the overall jumping efficiency should be about 50%.

Calculating from the figure of 46.7J/kg given in the article, the jumping efficiency is about 42% (=19.6/46.7 x 100%). In Table IV the efficiency is claimed to be about 80%, but this is probably relative to the maximum achievable efficiency of 50%, because 80% x 50% = 40%.

answered May 11, 2017 by sammy gerbil (26,096 points)
selected May 11, 2017 by gronk33
That makes complete sense. Now I understand. Thank you so much!!
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