What fraction of a 300-Calorie pizza slice must you eat to supply that energy. PLEASE HELP.
By
on
April 27th, 2009
You decide to remodel your kitchen and need to slide the refrigerator across the
horizontal floor over a distance of 5 m. The refrigerator weighs 400 lbs and slides with
constant speed. The coefficient of kinetic friction is 0.30.
a. As you push the refrigerator across the floor, how much work do you do.
b. What total energy do you expend. Assume 25% efficiency.
c. What fraction of a 300-Calorie pizza slice must you eat to supply that energy.
O-360 says:
a. The normal force is 400 lbs, or (400/2.2)*9.8 = 1782N. The friction force is μN = 1782*0.3 = 534N. The work done is W = F*d = 534*5 = 2,673J.
b. If the efficiency is 25%, you need to generate 4 times as much energy to delivery the necessary work.
Wt = 2,673/0.25 = 10,691J
c. Food “calories” are actually Kilo-Calories (Kcal), and 1Kcal = 4186 Joules. So, the 300 calorie slice of pizza is actually 300*4186 or 1.256*10^6 joules. It would take:
slice fraction = 10,691/1.256*10^6 = .0085 slice
April 27th, 2009 at 9:05 pm
Tupher says:
haha physics 131 umass. i don’t think you take into account the normal force because the box is only moving in the x direction. instead work equals only friction force i believe
April 27th, 2009 at 9:51 pm