RII-1. The free-body diagram for a car rounding a banked curve is shown. The car is going so fast that it is about to slip. The coordinate system has been chosen.What is the correct Y-equation:A: N sin + sN cos – mg = 0 B: N – sN sin – mg cos = 0C: N – mg cos = 0 D: N cos – sN sin – mg = 0E: None of these/ don't know.Answer: N cos – sN sin – mg = 0See web lecture notes, Review Exam II, page 1.RII-2. A rider in a "barrel of fun" finds herself stuck with her back to the wall. The free-body diagram is shown.Is the following statement definitely true for this situation? sN=mgA: Yes, it is certainly true. B: No, it might not be true. a mg sN N y x Answer: NO! It might not be true. The expression sN is the maximum possible force of static friction. The force of static friction can be any values less than this maximum: Fstatfric sN. If the barrel is turning so slowly that the rider is just about to slip down the wall, then Fstatfric = sN. If the barrel is turning very rapidly, then N is very large (since N=mv2/R) and sN > mg. Her weight mg is a constant, independent of the speed of rotation.RII-3. The tip of the nose of a pogo stick rider moves along the path shown. The maximum compression of the pogo stick spring is shown.At what point(A, B, C or D) is the elastic potential of the spring energy a maximum?Answer: D, when the spring is fully compressedAt what point is the gravitational potential energy a maximum?Answer: A, at the maximum height.At what point is the kinetic energy a maximum?Answer: C, which is when the bottom tip of the stick has just touched the ground and the spring is just starting to compress.ABCDxRII-4. A projectile is fired straight up and then it comes back down to its original height. There is air resistance as the projectile is moving. During the flight of the projectile, the work done by the force of gravity is..A: zero. B: positive C: negativeAnswer: Zero. The work done by gravity is negative on the way up, positive on the way down. The total work done is zero.During the flight of the projectile, the work done by the force of air resistance is...A: zero. B: positive C: negativeAnswer: Negative. The force of friction always opposes the motion, so the work done byfriction is negative on the way up and on the way down.During the flight of the projectile, the work done by the net force is...A: zero. B: positive C: negativeAnswer: Negative. Two ways to see this: 1. Work done by net force = Work done by gravity and by friction, so Wnet = Wgrav+Wfric. Wgrav=0 and Wfric<0 (from above questions) so Wnet<0.2. Appeal to the Work-Energy Theorem: Wnet = KE. Due to air resistance, the final speed of the projectile, just before it hits the ground is less than the initial upward velocity. Hence, the KE decreased and KE < 0. Therefore, Wnet < 0.RII-5. A new "constant force spring" is invented which has the remarkable property that the force exerted by the spring is independent of the stretch of the spring, Fspr = -q, whereq is a constant. What is the potential energy contained in this spring when it is stretched alength x.A: (1/2)qx2 B: q x C: q D: None of these.Answer: PE = qx. Apply the definition of PE: PE = –WField = +Wext = force distance =q x.This is just like our expression for gravitational potential energy near the Earth's surface. (ignoring signs) PE = Work = Force distance = mg h .RII-6 A planet is in elliptical orbit around the Sun. The zero of potential energy U is chosen at r = , so U rGMmr( ) . How does the magnitude of U (= -U) compare to the KE?A: -U > KE B: -U < KE C: -U=KED: depends on the position in the orbit.Answer: -U > KE When the velocity of the planet is the escape velocity, the total energy Etot=0, since the condition for escape velocity is KE PE KE PEmvGMmREi i f fesc tot 1220 0 0When v=vesc, Etot=0. When v>vesc, Etot>0. When v<vesc, Etot<0. If Etot is greater than or equal to zero, the planetcannot be in a bound orbit; it will escape toinfinity. Only when the Etot is negative is theplanet in a bound orbit. In order for Etot to benegative, the negative U must be larger inmagnitude than the positive KE. |U|>KE or-U>KE.U r Etot KE |U|RII-7. The gravitational potential energy of a rock near a star is shown in the diagram. When the rock is at the position shown, it has a kinetic energy of 25MJ. Will the rock escape to infinity or is it bound in orbit about the star?A: Escape B: in bound orbit C: impossible to tellAnswer: the rock is in a bound orbit. The total energy is Etot = PE + KE = -30MJ + 25MJ = -5MJ. The rock cannot escape the "gravity well" The maximum possible r the rock can possibly attain is where the Etot line intersects the U(r) curve. U(r) r -10MJ 0 -20MJ -40MJ U(r) r -10MJ 0 -20MJ -40MJ Etot PE(-) KE(+)CTRII-8. A pendulum is launched in two different ways. During both launches, the bob is given an initial speed of 3.0 m/s and the same initial angle from the vertical. On launch 1, the speed is upwards, on launch 2, the speed is downwards.Which launch will cause the pendulum to swing the largest angle from the equilibrium position on the left side?A: Launch 1 B: Launch 2C: Both launches give the same maximum displacement.Answer: Both launches give the same maximum displacement. Apply conservation of energy, Etot = KEi + PEi = KEmax = PEmax. For both launch 1 and 2, the initial KE and the initial PE are the same. So the Etot is the same for both launches.1 2CT9-9. A block of mass m with initial speed v slides up a frictionless ramp of height h inclined at an angle as shown. Assume no friction.True A or False B: Whether the block makes it to the top of the ramp depends on the mass of block and on the angle .Answer: false. The block will make it to the top if (1/2)mv2 > or = mgh. The m’s cancel.Whether the block makes it to the top depends only on v and h.Suppose now that there is friction between the block and the ramp surface.True A or False B: Whether the block makes it to the top of the ramp depends both on the mass of block and on the angle .Answer: False. Whether the block makes it to the top depends on the angle but not on the mass m. To see this, write out conservation of energy equation. Suppose the block slides up a distance x along the ramp before stopping, then we can writeinitial final212212(heat)cosi fkkE
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