Posted by ekaterina on .
A 68 student consumes 2500 Cal each day and stays the same weight. One day, he eats 3500 Cal and, wanting to keep from gaining weight, decides to "work off" the excess by jumping up and down. With each jump, he accelerates to a speed of 3.4 before leaving the ground.
Kinetic energy of jump = (1/2) m v^2
= (1/2)(68)(3.4)^2 = 393 Joules
.25(3500-2500) = 250 food calories to burn
1 food calorie (1000 physics heat calories) = 4184 Joules
250 Cal * 4184 J/Cal = 1,046,000 Joules
1,046,000/393 = 2661 Jumps
Luckily saw whole problem before the 25% efficiency part was deleted -
I was working on your final version.
Sorry, Damon. I thought she'd posted exactly the same question a dozen times, so I deleted the extras.
LOL - does not matter. I had already copied the last one.