Question

You overindulged on a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm raises holding a 5.00 kg weight in one hand. The distance from your elbow to the weight is 35.0 cm , and in each arm raise you start with your arm horizontal and pivot it until it is vertical. Assume that the weight of your arm is small enough compared with the weight you are lifting that you can ignore it. As is typical, your muscles are 20.0 % in converting the food energy they use up into mechanical energy, with the rest going into heat.
a. If your dessert contained 350 food calories, how many arm raises must you do to work off these calories?

Answer 1

Answer:

17086

Explanation:

We are given that

Weight=5 kg

Distance of elbow from the weight=d=35 cm=$\frac{35}{100}=0.35 m$

1 m=100 cm

g=$9.8m/s^2$

Work done  each time on the weight=$mgh=5\times 9.8\times 0.35=17.15 J$

20%  means 0.20 of food energy=Work done each time on weight

Total work done in each time=$\frac{17.15}{0.2}=85.75J$

1 food calorie=4186 J

350 food calories=$350\times 4186 J$

Number of lift needed=$\frac{350\times 4186}{85.75}=17086$

Hence, you must do 17086 arm raises to work off 350 food calories.