KE =? v o. Page 1 of 12

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1 Page 1 of 12 CTEnergy-1. A mass m is at the end of light (massless) rod of length R, the other end of which has a frictionless pivot so the rod can swing in a vertical plane. The rod is initially horizontal and the mass is pushed down with an initial speed v o. What initial kinetic energy is required for the mass to pivot 270 o to the vertical position? m R v o KE INITIAL = 1 2 mv o 2 A) mgr B) mg(2r) C) mg(3r) D) None of these Answer: mgr Begin by writing KE i + PE i = KE f + PE f. You get the same answer regardless of where you place the zero of PE. CTEnergy-2. A projectile is fired with an initial speed v o at an angle from the horizontal. What is the KE of the projectile when it is on the way down at a height h above the ground? (Assume no air resistance.) KE =? v o h A) (1/2)mv o 2 + mgh B) mgh C) (1/2)mv o 2 mgh D) Impossible to tell Answer: (1/2)mv o 2 mgh. Begin by writing KE i + PE i = KE f + PE f

2 Page 2 of 12 CTEnergy-3. A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times higher must the new ramp be? h v A) B) 2 C) 3 D) 4 E) none of these. Answer: 4 CTEnergy-4. 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. v v i = 0 While the mass is descending, its KE A) increases B) decreases C) remains constant. While the mass is descending, its PE A) increases B) decreases C) remains constant. While the mass is descending, its (KE + PE) = total mechanical energy A) increases B) decreases C) remains constant. Answers: The KE increases, the PE decreases, and the (KE+PE) decreases because some of the mechanical energy is converted into thermal energy.

3 Page 3 of 12 CTEnergy-5. What is the work done by the two forces together in moving the object a distance x to the right as shown in the diagram? The magnitude of each force is F and the directions of the two forces are 90 o apart. A) 2Fx B) Fx C) 2 Fx D) 1 2 Fx E) None of these. F F x Answer: Fx The vertical force does no work. CTEnergy-6. A mass is oscillating back and forth on a spring as shown. Position 0 is the unstretched position of the mass. E 0 M +E At which point is the magnitude of the force on the mass maximum? A) 0 B) M C) E At which point is the magnitude of the acceleration of the mass maximum? A) 0 B) M C) E At which point is the elastic potential energy maximum? A) 0 B) M C) E Answers: force is max at E, acceleration is max at E, PE is max at E. CTEnergy-7. A spring-loaded dart gun shoots a dart straight up into the air, and the dart reaches a maximum height of h. The same dart is shot straight up a second time from the same gun, but this time the spring is compressed twice as much before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring? A) 2h B) 4h C) 8h D) h/4 E) h/2 Answer: 4h

4 Page 4 of 12 CTEnergy-8. A spring-loaded gun fires a dart at an unknown angle above the horizontal. No air resistance. The spring constant k and the initial compression x of the spring are known. Can you solve for the maximum height h above the starting point using Conservation of Energy? A)Yes B) No k, x h =? Answer: No. The KE at the top of the trajectory is unknown, so you can t solve for h in terms of known quantities. CTEnergy-9. 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? v = 4m/s h = 1m A) Yes B) No C) impossible to determine from the info given Answer: No. The initial KE is (1/2)mv 2 = 8m (in SI units). The PE at the top is mgh = 9.8m (in SI units). There is not enough initial KE to convert into PE.

5 Page 5 of 12 CTEnergy-10. A small mass, starting at rest, slides without friction down a rail to a loop-de-loop as shown. The maximum height of the loop is the same as the initial height of the mass. Will the ball make it to the top of the loop? A) Yes, the ball will make it to the top of the loop. B) No, the ball will not make it to the top. C) Not enough info to say Answer: This was a tough one! No, the ball will not make it to the top. The ball needs to have significant KE at the top of the loop in order to remain in contact with the rail. If, instead of an open rail, the ball was in a tube, as shown below, then the ball would make it to the top. Same question, but now the loop-de-loop track is replaced by a frictionless tube. Answer: Yes, now the ball will make it to the top.

6 Page 6 of 12 CTEnergy-11. A cart rolls without friction along a track. The graph of PE vs. position is shown. The total mechanical energy (KE + PE) is 45kJ. The top of the hill at x = 34 m is round, and the bottom of the valley near x = 55 m is flat. PE (kj) E tot x(m) To within 5 kj, what is the maximum KE over the stretch of track shown? A) 55kJ B) 10kJ C) 45kJ D) 35 kj E) None of these Answer: 35 kj When the KE is a minimum (over this stretch of track), what is the direction of the acceleration? A) up B) down C) right D) Left E) some other direction (at an angle) or zero. Answer: down. When the KE is min, it is on the top of the round hill. For circular motion with constant speed, the acceleration is toward the center of the circle. When the KE is max, what is the direction of the acceleration? A) up B) down C) right D) Left E) some other direction (at an angle) or zero. Answer: no direction, the acceleration is zero. When the KE is max, it is in the flatbottomed valley, moving with constant velocity.

7 Page 7 of 12 CTEnergy-12. A cart rolls without friction along a track. The graph of PE vs. position is shown. The total mechanical energy (KE + PE) is 0 kj. 10 PE (kj) E tot x(m) What is the MAX KE of the cart during its journey (to within 5kJ)? A) 30 kj B) 50 kj C) -30 kj D) 10 kj E) None of these Answer: 30 kj Suppose the cart is at position x = 20 m, is moving right, and has total energy E tot = 15 kj. Will the cart make it over the hill at x=34 m? A) Yes. B) No. Answer: No. It need a total energy of at least 10 kj to make it over the hill.

8 Page 8 of 12 CTEnergy-13. Suppose an unusual (non-hooke's Law) spring has a force which is a constant, independent of positive stretch length x : F spring = C, where C is a positive constant. (Instead of F sping = kx). The potential energy associated with this spring is (for positive stretch x) is A) PE C B) PE Cx C) PE 1 C x 2 2 D) PE 1 C x 3 3 E) None of these Answer: PE C x Remember the definition of PE: PE F = W F = +W ext CTEnergy-14. 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 P 1. Elevator 2 can carry the same load up twice the distance (2h) in twice the time (2t). What is its power output P 2? A) P 2 = P 1 B) P 2 = 2P 1 C) P 2 = 4P 1 D) None these Answer: P 2 = P 1 P = W/ t = mgh/t 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 P 3? A) P 3 = P 1 B) P 3 = 2P 1 C) P 3 = 4P 1 D) None these Answer: P 3 = 4P 1 h time t m m 2m 1 2h time 2t 2 3 2h time t

9 Page 9 of 12 CTEnergy-15. Student A climbs a flight of stairs 10 m high in 8 s. Student B climbs a flight 20 m high in 16 s. Which student is producing the greater mechanical power output? A) A B) B C) Tie 16 s 8 s 20 m 10 m Answer: Tie. The mechanical power output is the power that is going into increasing the gravitational PE = mgh. Power = mgh/ t

10 Page 10 of 12 CTEnergy-16. 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) 1.5 m B) 2 m C) 3 m D) 4 m 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 (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

11 Page 11 of 12 CTWork-17. A system consists of a mass m hanging from a spring (spring constant k) and the Earth as shown. In situation A, the mass is hanging freely, at rest, so the spring is stretched beyond its relaxed length. In situation B, the mass is also stationary, but the spring is now at its relaxed length, because an alien from Planet X has lifted the mass and is holding it up. A B In which situation is the total energy of the mass/spring/earth system bigger? A) A B) B C) same energy in both. Answer: B. The Alien visitor did work on the system when he lifted it, and this increased the energy of the system. The energy of a system is only constant if the system is isolated.

12 Page 12 of 12 CTWork-18. Three equal masses are tied together (massless strings) and pulled to the left by a force of magnitude F over a frictionless, horizontal surface. The free-body diagrams for the 3 masses are shown. Some of the 11 forces shown in the free-body diagrams are Newton s 3 rd Law Action- Reaction pairs. How many action-reaction pairs can you count among the 11 forces shown? F A) 2 B) 3 C) 4 D) 5 E) None of these Answer: 2 pairs