Energy Transfers: Work, Potential Energy, and Kinetic Energy, Study notes of Physics

Download Energy Transfers: Work, Potential Energy, and Kinetic Energy and more Study notes Physics in PDF only on Docsity! CTEnergy-1 Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by … the force of gravity is.. A) + B)  C) 0 the force of Albert's hand is.. A) + B)  C) 0 the net force on the book is .. A) + B)  C) 0 Answers: The work done by gravity is + since displacement and force of gravity are in the same direction. The work done by Al's hand is (–), since displacement and the force of Al's hand are in the opposite direction. The net force is zero, so the work done by the net force is zero. CTEnergy-2 A rock of mass m is twirled on a string in a horizontal plane. The work done by the tension in the string on the rock is.. A) + B)  C) 0 Answer: The work done by the tension force is zero, since in any tiny interval of time, the force of the tension in the string is perpendicular to the direction of the displacement. s F s F hand F grav CTEnergy-3 A 1 kg mass is moved part way around a square loop as shown. The square is 1 meter on a side and the final position of the mass is 0.5 m below its original position. Assume that g = 10 m/s2. What is the work done by the force of gravity during this journey? mgh = (1 kg) (10 m/s2) (h) A) 10 J B) 5 J C) 0 J D) -10 J E) -5 J Answer: Total work done is +5 J. The work done during the horizontal segments is zero (since force is perpendicular to displacement there). The work during the first upward vertical segment is negative and this negative work is cancelled by the positive work done during the first half of the downward vertical segment. The total work is that done during the second half of the downward vertical segment where h = 0.5 m. If instead of moving part way around a square, the mass where taken on a long and tortuous journey to the Moon, Tibet, and Lithuania and then returned to the same finish point as before, would the work done by gravity be the .. A) same, or B) different. Answer: the same! Any journey can be thought of a series of small vertical or horizontal displacements. During any horizontal segment, the work done by gravity is zero. All upward vertical segments are cancelled by corresponding downward vertical segments, EXCEPT for the last 0.5 m between the start and the finish. CTEnergy-4 Two marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. (Assume no air resistance.) Just before hitting the ground, the heavier marble has.. A) as much kinetic energy as the lighter one. B) twice as much kinetic energy as the lighter one. C) half as much kinetic energy as the lighter one. D) four times as much kinetic energy as the lighter one. E) impossible to determine. start finish 1m 0.5m Up In which case, we get the same answer as before: i i f f i i KE PE KE PE KE mgR 0 mg2R KE mgR        CTEnergy-7 A mass slides down a rough ramp (with friction) of height h. Its initial speed is zero. Its final speed at the bottom of the ramp is v. As the mass descended, its KE A) increased B) decreased C) remained constant. Answer: KE increased As the mass descended, its PE A) increased B) decreased C) remained constant. Answer: PE decreased As the mass descended, its (KE + PE) = total mechanical energy A) increased B) decreased C) remained constant. Answer: Emech decreased. Some of the mechanical energy was converted into thermal energy by the friction. Remember: KE+PE = constant ONLY if NO friction and no heat generated. y = +R y = 0 y = +2R v vi = 0 CTEnergy-8 A hockey puck slides without friction along a frozen lake toward an ice ramp and plateau as shown. The speed of the puck is 4m/s and the height of the plateau is 1m. Will the puck make it all the way up the ramp? A) Yes B) No C) impossible to determine without knowing the mass of the puck. Answer: We have to do a quick calculation. Setting y = 0 at the bottom of the ramp, we have.. i i f f 21 2 KE PE KE PE mv 0 0 mgR       The final energy (PE) would be mgR = mass(10m/s2)(1m) = mass10 m2/s2 The initial energy (KE) is (1/2)mv2 = mass(0.5)(16 m2/s2) = mass8 m2/s2 So the initial KE is not large enough; the puck will not make it to the top of the hill. CTEnergy-9 A mass is oscillating back and forth on a spring as shown. Position 0 is the relaxed (unstretched) position of the mass. 0 M E At which position is the magnitude of the acceleration of the mass a maximum? A) 0 B) M C) E At which point is the magnitude of the force on the mass a maximum? A) 0 B) M C) E At which point is the elastic potential energy a maximum? A) 0 B) M C) E Answers: The acceleration is maximum at E, since the force Fspring = –kx is maximum magnitude there. (Newton's second law says Fnet = m a, so the acceleration is max when Fnet is max.) h = 1m v = 4m/s The force on the mass is maximum at E (for the same reason that acceleration is max there.) The elastic potential energy (PEelas = ½ kx2) is maximum at E (where x is max). Ch6-10 A spring-loaded dart gun shoots a dart straight up into the air, and the dart reaches a maximum height of h = 24 m. The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring? A) 2h = 48m B) h = 24m C) h/2 = 12m D) h/4 = 6m E) h/8 = 3m Answer: Use conservation of energy and write Ei = Ef with the initial position = where the dart is inside the gun, with the spring cocked, and the final position = top of the trajectory. Let's set h = 0 at the initial position. i grav,i elas,i f grav,f elas,f 21 2 21 2 2 2 KE PE PE KE PE PE 0 0 k x 0 m g h 0 k x mg h x h , x constant h C h                 Here C is a constant (C = 2mg/k). Since x2 is proportional to h, when x is cut in half [ x  x/2 , x2  (x/2)2 = x2/4 ] , h is reduced by a factor of 4. It may be easier to see this by setting up a ratio: C 2h C    2 22 1 1 12 12 4 1 42 2 2 1 1 11 x xx x x xh      A spring-loaded dart gun shoots a dart straight up into the air, and the dart reaches a maximum height of h = 24 m. The same gun is reloaded with the spring compressed the same amount, but now the gun is aimed at an angle of 45o to the horizontal. Will the dart reach the same maximum height of 24m? Assume no frictional losses. A) Yes, the dart will reach the same height B) No, the dart will not reach the same height. Answer: No. We can see why this must be, by considering conservation of energy. When fired at 45o, the dart has some horizontal motion when it is at the top of it trajectory. So it always has some KE, and its initial PEelastic is never fully converted into PEgrav. The total mechanical energy (KE + PE) is 0 kJ. What is the MAX KE of the cart during its journey (to within 5kJ)? A) 35 kJ B) 48 kJ C) 16 kJ D) -16 kJ E) None of these/don't know Answer: Max KE is where PE is min. Max KE = +48 kJ Suppose the cart is at position x = 20m, is moving right, and has total energy Etot = –20kJ. Will the cart make it over the hill at x=38m? A) Yes. B) No. Answer: No. The "turning point" = where the cart slows to a stop and turns around. The turning point is where the Etot line intersects the PE curve. This is where KE = 0 because Etot = KE + PE. PE = –48kJ KE = +48kJ E tot = 0 0kJ -20kJ -40kJ -60kJ 20m 60m 40m CTEnergy-15 A hockey puck sliding on an ice rink is moving at 1 m/s when it slides onto a carpet that someone left on the ice. The puck comes to rest after moving 1m on the carpet. How far along the carpet would the puck go, if its initial speed was 2m/s? A) 1.5m B) 2m C) 3m D) 4m E) Impossible to determine. [Hint: Apply the Work-Energy Theorem. If the puck slides twice as far, the friction does twice as much (negative) work.] Answer: 4 m. We have to do a calculation to see this. This is a case with friction. Remember that the magnitude of the work done by friction is the amount of thermal energy gained, which is the amount of mechanical energy (KE+PE) lost. fric 21 K 2 2 W KE (since PE = 0) mg x m v x v         So if we double v, v2 is increased by a factor of 4, and so is x. Warm-up question) How much more KE does the puck have moving at 2 m/s compared to moving at 1 m/s? A) 2 times as much B) twice as much C) 3 times as much D) 4 times as much Answer: 4 times as much E tot = –20 kJ CTEnergy-16 Elevator 1 can carry a load of mass m up a distance h in a time t, before the engine overheats. Its power output is P1. Elevator 2 can carry the same load up twice the distance (2h) in twice the time (2t). What is its power output P2 ? A) P2 = P1 B) P2 = 2P1 C) P2 = 4P1 D) None these Answer: P2 = work/time = mg(2h)/2t = mgh/t = P1 CTEnergy-17 Elevator 3 can carry twice the load (2m) up twice the distance (2h) in the same time (t) as elevator 1. What is its power output P3 ? A) P3 = P1 B) P3 = 2P1 C) P3 = 4P1 D) None these Answer: P3 = work/time = (2m)g(2h)/t = 4mgh/t = 4P1 1 h time t 2 3 2h time 2t 2h time t m m 2m