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1) Each of the boxes are pulled 10 m across a level, frictionless floor by a for

ID: 1477730 • Letter: 1

Question

1) Each of the boxes are pulled 10 m across a level, frictionless floor by a force given. Which bod experiences the greatesst change in its kinetic energy?

10 N with 1.0 kg mass

15 N with 2.0 kg mass

5.0 N with 0.50 kg mass

20 N with 3.0 kg mass

12 N with 0.30 kg mass

2) At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made up by an equivalent food intake, how many kilocalories (food calories) does this person need to consume every day just to make up this heat loss? (1 cal= 4.186 J)

3) What is the best possible performance coefficient (COP) of a heat pump which is used to cool down a fridge to a temperature of 280 K if the room temperature is 292 K?

4) The weight of a 1200 kg car is supported equally by the four tires, which are inflated to the same gauge pressure. What gauge pressure is required so the area of contact of each tire with the road is 100cm^2? (100cm^2=0.01m^2)

5) How many air molecules are in a balloon? (V=0.07 m^2, T=292 K, p=130 kPa)

Explanation / Answer

using work energfy theorem

work done by the net force = change in KE

F*S = dKE

For 10 N with 1.0 kg mass

change in KE = dKE = F*S = 10*10 =100 J

For 15 N with 2.0 kg mass


dKE = 15*10 = 150 J

5.0 N with 0.50 kg mass

dKE = 5*10 = 50 J

20 N with 3.0 kg mass

dKE = 20*10 = 200 J

12 N with 0.30 kg mass


dKE = 12*10 = 120 J

So greatest change in kinetic energy is 20 N with 3.0 kg mass

2) heat loss = power *time = 62*24*60*60 = 5356800 J

but 1 cal = 4.186 J

1J = 1/4.186 cal = 0.238 cal


then required energy consume is 5356800*0.238 = 1279694.21 cal = 1279.69 kcal

3) COP = Th/(Th-Tc) = 292/(292-280) =24.33


4) guage pressure is P = F/A = (1200*9.81/4)/(100*10^-4) = 294300 Pa


5) using P*V = n*R*T

130000*0.07 = n*8.314*292


no.of moles n = 3.74 moles

the no.of air molecules are 3.74*6.022*10^23 = 2.257*10^24 molecules