Chapter 3 Problem Set 2. How much work is needed to raise a 100-kg load of bricks 12 m above the ground to a building under construction? Answer: The relevant formulas here are: W = F X d (work = force X distance) and F = m X a (force = mass X acceleration) In this case we are give the mass (110-kg), and the distance (12 m), and we know the acceleration due to gravity is 9.8 m/sec2. So we first solve for the force and then solve for the work done. F = m X a = 110 kg X 9.8 m/sec2 = 1,078 kg*m/sec2 = 1,078 N (newtons) Plugging that into the formula for work: W = F X d = 1,078 N X 12 m = 12,936 N*m = 12,936 J (joules) 10. What is the speed of an 800-kg car whose KE (kinetic energy) is 250 kJ? The relevant formula here is: KE = ½ mv2 (kinetic energy = ½ X mass X velocity2 In this case we are give the mass (800-kg) and the kinetic energy (250 kJ) and we are asked to solve for the speed of the car. So first we convert kilojoules (kJ) into joules : 250 kJ = 250 kJ X 1000 J/kJ = 250,000J Rearanging the formula for kinetic energy to solve for velocity we get: NOTE: The symbol followed by brackets [ ] will be used to show the square root function for all problems. So the square root of 25 would be shown as [25]. V2 = [2 X (KE)/m] v = [(2 X 250000 J)/800kg] = [(2 X 250000 N*m)/800 kg] = [(2 X 250000 kg*m2/sec2)/800 kg] = [(500,000 k*m2/sec2)/800 kg] = [625 m2/sec2] = 25 m/sec 15. A 70-kg athlete runs up the stairs from the ground floor of the Empire State Building to its one hundred second floor, a height of 370 m, in 25 min. How much power did the athlete develop? The relevant formulas here are: F = m X a (force = mass X acceleration; here the acceleration is that generated by gravity g) W = F X d (work = force X distance), P = W/t (power = work/time) We are given the athlete’s mass (70-kg), and the distance involved is the height that he runs to (370 m) and we know that the acceleration due to gravity is 9.8 m/sec2. We are also give the amount of time required to do the amount of work the athlete did (25 min). First, solving for the force involved: F = mg = 70 kg X 9.8 m/sec2 = 686 kg*m/sec2 = 686 N (newtons) Now solving for the work done: W = F X d = mgh = 686 N X 370 m = 253,820 N*m = 253,820 J (joules) Before we can solve for power we have to convert the time (25 min) into seconds: t = 25 min X 60 sec/min = 1,500 sec Now solving for the power: P = W/t = 253,820 J/1500 sec = 169 w (watts) 24. A boy throws a 4-kg pumpkin at 8 m/sec to a 40-kg girl on roller skates, who catches it. At what speed does the girl then move backward? This is a case of the law of the conservation of momentum. The relevant formula is: p = m X v (momentum = mass X velocity) We are given the mass (4-kg) and velocity (8 m/sec) of the pumpkin so we can calculate its momentum (p). We are also given the weight of the girl (40-kg). Since momentum is conserved, we can rearrange the formula (p = m X v) and calculate for the speed of the girl. Solving for the momentum of the pumpkin: p = m X v = 4 kg X 8 m/sec = 32 (kg*m)/sec Rearranging the equation to solve for the girl’s velocity: v = p/m = (velocity = momentum/mass) = [32 (kg*m)/sec]/40 kg = 0.8 m/s 31. Approximately 5.4 X 106 J of chemical energy is released when 1 kg of dynamite explodes. What fraction of the total energy of the dynamite is this? The relevant formula here is: E0 = mc2 [resting energy = mass X (speed of light)2] We are given the mass of the dynamite (1-kg), the chemical energy (Ec) given off by the explosion (5.4 X 106 J) and we know the speed of light (3 X 108 m/sec) so we can calculate the total resting energy (E0) of the dynamite: E0 = mc2 = 1 kg X (3 X 108 m/sec)2 = 1 kg X (9 X 1016 m2/sec2) = 9 X 1016 kg*m2/sec2 = 9 X 1016 J (joules) The fraction of the total energy given off as chemical energy is a ratio of the two: Ef = Ec/E0 (fractional energy = chemical energy/total resting energy) Ef = (5.4 X 106 J)/(9 X 1016 J) = 6 X 10-11 or 6 X 10-9 % (0.000000006 %)