Homework 3 solutions

Useful equations for HW3:

1	KE = 1/2mv2	The kinetic energy of a moving object is equal to one half of its mass times its velocity squared.
2	PE = mgh	The potential energy of an object in a uniform gravitational field is equal to the mass of the object times the local acceleration due to gravity (9.8m/s2 at Earth's surface) times the height.
3	P = W/t	The power output of a system is equal to the amount of work done by the system divided by the time over which that work was exerted.
4	W = fd	The work done by a force acting on an object is equal to the product of the force times the distance over which the force is applied
5	deltaQ = deltaTHcm	The heat necessary to change the temperature of a system by deltaT is equal to the change in temperature times the heat capacity times the mass of the object being heated.

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	PROBLEM 1: Georgie was pulling her brother (20kg) in a 10-kg sled with a constant force of 25N for one block (100m). a. What is the work done by Georgie? b. How long would a 100W light bulb have to glow to produce the same amount of energy exerted by Georgie? a. The work done by the puller we can obtain directly from equation 4 above, that the work is equal to the force times the distance. W = Fd W = 25N * 100m
1/2 point	W = 2500 N-m
	b. This part of the problem asks for how long (a time) a light would have to be on to have exert the same amount of energy as the puller. In this instance the amount of energy exerted is equal to the work, and we can set that equal to the amount of energy emitted by the light bulb. The light bulb's energy we can obtain from equation 3 since we know the power output. P = W/t W = Pt Elightbulb = W = Pt EGeorgie = W = Fd = 2500 N-m This can get a bit tricky here in that we have to be sure that our units work out correctly. A Newton is a kg-m/s2. A Watt is a Joule per second = J/s. A Joule is a kg-m2/s2, which is equivalent to a N-m (Newton meter). Elightbulb = EGeorgie Pt = 2500 N-m = 2500 J t = 2500 J / P t = 2500 J / 100W = 2500 J / 100J/s
1/2 point	t = 25 s
	So it turns out that the mass that it gave us is irrelevant to the question asked.
	PROBLEM 2 You throw a softball (250g) straight up into the air, it reaches a maximum altitude of 15m, and then it returns to you. (Assume that the ball departed and returned at ground level.) a. What is the gravitational potential energy (in joules) of the softball at its highest position? b. What is the kinetic energy of the softball as soon as it leaves your hand? (Assume no energy loss by the softball while it is in the air) c. What is the kinetic energy of the softball when it returns to your hand? d. From the kinetic energy, calculate the velocity of the ball as it left your hand. a. The gravitational potential energy calculation is a straightforward implimentation of equation 2, equal to the mass times the gravity times the height. PE = mgh PE = 250g * 9.8 m/s2 * 15 m The problem specifically asks for the answer in Joules, and since a joule is 1 kg*m2/s2 we need to convert grams to kilograms. PE = .250kg * 9.8 m/s2 * 15 m
1/4 point	PE = 37 J
	b. The kinetic energy of the ball as it leaves your hand (KEi) is exactly equal to the potential energy of the ball at the height of its trajectory. It has transferred the kinetic energy you imparted to it entirely to potential energy (its not moving at the very top of the arc so its KE is 0).
1/4 point	KEi = 37 J
	c. Since no energy has been lost and the height is 0 when it returns to your hand, all of the potential energy has been converted back into kinetic energy so the kinetic energy upon return (KEf) is also equal to the potential energy at the maximum.
1/4 point	KEf = 37 J
	d. The kinetic energy of the ball when it returns to your hand is equal to 1/2mv2 from equation 1 above. Set these equal and solve for v: KE = 1/2mv2 2*KE = mv2 2*KE/m = v2 v = (2*KE/m)1/2 v = (2 * 36.75 J * / .25kg)1/2 v = (294 J/kg)1/2 = (294 kg*m2/s2 / kg)1/2
1/4 point	v = 17 m/s
	PROBLEM 3 While skiing in Jackson, Wyoming, your friend (65kg) starts his descent down the bunny run, 25m above the bottom of the run. Assume he starts at rest and converts all of his gravitational potential energy into kinetic energy. a. What would be your friend's kinetic energy at the bottom of the bunny run? b. What would be his final velocity? c. Is this speed "reasonable"? a. The kinetic energy in the bottom will be the same as the kinetic energy at the top, just as the question says. KE = PE = mgh KE = 65kg * 9.8m/s2 * 25m KE = 15,900 kg m2/s2
1/3 point	KE = 1.6 x 104 J
	b. For the final velocity just set the expression for kinetic energy (equation 1) equal to the KE we just determined in part (a). KE = 1/2mv2 v = (2*KE/m)1/2 v = (2 * 15900 J / 65kg)1/2 v = (490 m2/s2)1/2
1/3 point	v = 22 m/s
	c. Hmmm. 22 m/s seems pretty fast, lets convert it to something I understand a bit better, miles per hour. 22 m 100 cm 1 in inches foot mile 3600s s m 2.54cm 12 inches 5280 feet hour
1/3 point	Grinding this out on the calculator we get 50 mi/hr. That's pretty fast, but olympic skiiers go even faster. Both answers ("yes" and "no") were given full credit, as long as you converted it to mi/hr or something you could more easily comprehend.
	PROBLEM 4 Outline the three major modes of heat transfer. In this outline, state if a medium is necessary, compare the motion of molecules in this medium, and relate the heat transfer to the presence or absence of a temperature difference in the medium.
all parts worth 1/12 point	I. Conduction a. Medium not required, though can be used. b. Atoms in hot body pass heat to the cold body through the kinetic energy of the intervening atoms. b. Only occurs with hot to cold temperature differential. II. Convection a. Medium required. b. Hot atoms are moved en masse to a cold area and vice versa. c. Temperature difference required, but flows both directions at once.. III. Radiation a. No medium required. b. Atoms radiate according to their temperature (blackbody radiation). c. No temperature difference required, all atoms above absolute zero radiate.
	PROBLEM 5 The heat capacity of the average human body is 83% of the heat capacity of water. How much energy would it take to raise the temperature of a 60kg human being by 1oF? By 1oC? By 1 Kelvin? The heat capacity of water is 1 calorie / goC. If the heat capacity of a human is 83% of this than the heat capacity of a human body is: Hc human = .83 * Hc water Hc human = .83 * 1 calorie/goC Hc human = .83 calorie/goC which we'll refer to as Hc from here on. Applying equation 5 (be careful of your units!): Q = THcm Q = 1oC * .83 calorie/goC * 60kg Q = .83 calorie/g * 60kg = .83 calorie/g * 60000g
1/3 point	Q = 50,000 calories
	which is 50 of the Calories you see on the back of food packages. Since 1oF is 5/9 of a oC the answer for raising the temp by 1oF is 5/9 of the one we just obtained or.
1/3 point	28,000 calories
	One Kelvin is equal in size to one oC so the answer then for one K is the same as for one oC.
1/3 point	50,000 calories