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.

Question

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.

Damon · Answer

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

so
1,046,000/393 = 2661 Jumps

Damon · Answer

I was working on your final version.

Ms. Sue · Answer

Sorry, Damon. I thought she'd posted exactly the same question a dozen times, so I deleted the extras.

Damon · Answer

LOL - does not matter. I had already copied the last one.