This test covers Work, echanical energy, kinetic energy, potential energy (gravitational and elastic), Hooke s Law, Conservation of Energy, heat energy, conservative and non-conservative forces, with soe probles requiring a knowledge of basic calculus. Part I. Multiple Choice F Ø. A force F is exerted at an angle Ø on a box of ass as it is dragged across the floor at constant velocity. If the box travels a distance x, then the work done by the force F on the box is a. Fx b. Fx cos Ø c. gx cos Ø d. Fx sin Ø e. Fx tan Ø 2. A block of wood, initially oving along a rough surface, is pushed with an applied horizontal force F applied that is less than the friction force F friction. Which of the following stateents is false? a. The Work being done by the applied force is negative. b. The net Work being done on the block is negative. c. The block is slowing down. d. The net Work being done on the box decreases its kinetic energy K. e. There is an increase in internal energy due to friction. 3. A 300-Watt electric wheelchair has a ass of 50kg, and carries its 50kg occupant at constant velocity up a long rap. About how uch tie does it take the wheelchair to reach the top of the 0-eter high rap? a. 3 s b. 7 s c. 0 s d. 333 s e. 33 s
U(r) 3U0 2U0 U0 0 r0 2r0 3r0 4r0 r 4. The graph above represents the potential energy U as a function of position r for a particle of ass. If the particle is released fro rest at position r o, what will its speed be at position 3r 0? a. 8U 0 b. c. 4U 0 2U 0 d. 6U 0 e. The particle will never reach position 3r 0. 5. The behavior of a non-linear spring is described by the relationship F = 2kx 3, where x is the displaceent fro the equlibriu position and F is the force exerted by the spring. How uch potential energy is stored in the spring when it is displaced a distance x fro equilibriu? a. 2 kx 4 b. c. d. e. 6kx 2 3 kx 4 3 kx 3 2 3 kx 2
Part II. Free Response anchor 0 d 2d anchor 0 6. A block of ass rests on a rough surface, and has a light spring of spring constant k and unstretched length d attached to one side as shown, with the other end of the spring attached to an anchor. There is a static coefficient of friction µ s between the surface and the block, and when the block is placed to the right at position 2d, it reains stationary on the surface. Express answers in ters of, k, d, and fundaental constants. a. Draw a free-body diagra of the block at the tie when it is located at position 2d. d 2d b. Deterine the friction force acting on the block when it is located at position 2d.
anchor 0 d 2d 3d c. The block is now oved to position 3d and released, where it reains at rest. When the block is oved slightly past this position, the block begins to slide along the surface with a kinetic coefficient of friction µ k. i. In ters of the variables given, what is the value of µ s? ii. How uch potential energy is stored in the ass-spring syste just before the block begins to ove? iii. The block slides a total distance of d before coing to a halt again. Deterine the coefficient of kinetic friction µ k.
iv. At what position does the block have its axiu velocity as it slides? v. What is the axiu velocity of the block as it slides?
A r = 20 D h = 60 C E B 7. A roller coaster car of ass = 200 kg is released fro rest at the top of a 60 high hill (position A), and rolls with negligible friction down the hill, through a circular loop of radius 20 (positions B, C, and D), and along a horizontal track (to position E). a. What is the velocity of the car at position B? b. Deterine the velocity of the car at position C. c. Draw a free-body diagra of the car at position C.
d. Deterine the velocity of the car at position D. e. Deterine the force (agnitude and direction) of the track on the car at position D. f. After copleting the loop, the rollercoaster car is travelling horizontally at velocity v 0 and subjected to a braking force F braking = kv, where k is a constant, v is the instantaneous velocity of the car, and tie t is the aount of tie that the braking force has been applied. i. Develop a definite integral expressed in ters of initial velocity v 0, k,, and tie t that could be used to evaluate the velocity of the car along the horizontal track. ii. Solve the integral to deterine an equation that could be used to calculate the horizontal velocity of the car as a function of initial velocity v 0, k,, and tie t.
8. A conservative force acts in the x-direction on a particle of ass = 2.0 kg to produce a potential energy curve as shown above. a. A particle is released fro rest at the 0.5 eter position. i. What is the potential energy of the particle at this position? ii. What is the velocity of this particle at x = 2 eters? iii. Describe the point on the U curve at x = 4 briefly, and what happens when the released particle reaches this position. iv. Does the released particle reach x = 7? If not, explain why not. If so, describe the particle s behavior.
b. A second particle, also of ass, is released fro rest at x = 8. Briefly describe the behavior of this particle. c. Sketch a graph of the conservative force that produces this potential energy curve.
Practice Test Solutions:. The correct answer is b. Work done by an object is calculated according to the Work forula W=F x, or W=Fx cos Ø. There are a couple of distractors in this proble: the ass of the box is not needed in the solution, and the box s constant velocity isn t required. The fact that the box is traveling at constant velocity iplies that there is friction ipeding its otion, but evidently the energy lost to friction is equal to the work being done by the force F, so that the net Work done on the box by F and F friction = 0. None of this inforation is necessary, though, to solve the proble. 2. The correct answer is a. Stateent a is false because the work done by any single agent is W = Fx cosθ. Here, with the applied Force and the displaceent of the box being in the sae direction (or cosθ = cos(0) =), the Work being done by the applied force is positive, and goes toward increasing the box's kinetic energy. Of course, the net, or overall, Work being done on the box has to include the force of Friction, which is acting in a direction opposite that of the box's displaceent. Thus, Work done by friction is negative, which has the effect of converting soe of the box's kinetic energy to internal energy via heat. 3. The correct answer is e. The wheelchair carries a total ass of 00kg up to a height of 0, with 300J of Work being done by the wheelchair each second. The tie for the total Work does is calculated as follows: P = Work tie tie = Work Power = gh P tie = (00kg)(0 /s2 )(0) 300J /s = 0000 300 = 33s 4. The correct answer is c. This is a conservation of energy proble, in which we exaine the relationship between the particle s potential and kinetic energies. U i + K i = U f + K f 3U 0 + 0 = 2U 0 + v 2 2 U 0 = v 2 2 v = 2U 0 5. The correct answer is a. The potential energy stored in the spring is calculated using the Work integral: U = U = x f x i x 0 U = 2k x 4 F dx 2kx 3 dx x 4 0 = 2 kx 4
Practice Test Solutions: 6. a. The block isn t oving yet, so the force exerted by the spring to the left equals the force exerted by friction to the right. Vectors should be labeled and drawn with lengths proportional to their agnitude. Soe instructors wish for vectors to be drawn with their point of origin beginning at the location where that force is applied, and soe instructors prefer that force vectors be drawn originating fro the center of the (point) ass. FNoral Fspring Ffriction Fgravity b. The friction force acting on the block at this point is equal in agnitude to the force applied by the spring: F net = a c. F spring F friction = 0 F friction = kx = kd i. The coefficient of static friction is based on the axiu force of static friction that the blocksurface can support. In this case: µ static = F friction F Noral = kx g = k(2d) g ii. The potential energy stored in the spring is siply based on U spring = 2 kx 2 : U spring = 2 kx 2 U spring = 2 k(2d)2 = 2kd 2 iii. The coefficient of kinetic friction can be deterined by using a Conservation of Energy analysis, taking into account the potential energy of the ass-spring syste at the beginning and end, as well as the energy lost to heat in the block s slide: U spring initial ΔE internal = U spring final 2 kx 2 i F friction d = 2 kx 2 f 2 kx 2 i µ k gd = 2 kx 2 f 2 k(2d)2 µ k gd = 2 kd 2 µ k = 3kd 2g 20, Richard White www.crashwhite.co
Practice Test Solutions: Note that this result reveals that µ k = 3 " 2kd% $ ' = 3 4 # g & 4 µ x, which is consistent with the principal that coefficients of kinetic friction are less than coefficients of static friction. iv. The block has its axiu velocity where the slope of the velocity-tie curve is 0, ie. where acceleration is 0. At that position, the force fro the spring and the force of friction are equal to each other: F net = a F spring F friction = 0 kx µ k F Noral = kx µ k g = 0 where x is the displaceent of the spring fro its unstretched length at d. Substituting in µ k fro the proble before: kx µ k g = 0 " kx 3kd % $ 'g = 0 # 2g & kx 3 2 kd = 0 7. x = 3d, relative to unstretched length at d 2 Relative to the origin: 3d 2 + d = 5 2 d v. Now that we know the position of the axiu velocity, we can use Conservation of Energy again to calculate that velocity: U s initial ΔE int = K +U s final 2 kx 2 i F friction x = 2 v2 + 2 kx 2 f # 2 k(2d)2 µ k g d & % $ 2 ' ( = 2 v2 + 2 k # % 3 $ 2 d & ( ' # 2 k(2d)2 3kd & % $ 2g ' (g # d & % $ 2 ' ( = 2 v2 + 2 k # % 3 $ 2 d & ( ' Solve to get v = d 2 k a. This is a conservation of energy proble, with the gravitational potential energy of the car U being converted to kinetic energy K as the car rolls down the hill. U = K 2 2 gh = 2 v 2 v = 2gh = 2 9.8 60 = 34.3 /s
Practice Test Solutions: b. At position C, soe of the car s K has converted back to U. U i = K +U f gh A = 2 v 2 C + gh C v C = 2g(h A h C ) = 2 9.8(60 20) = 28 /s c. The free-body diagra for the car includes the only two forces acting on the car: the force of gravity, and the force of the track which is providing the centripetal force that keeps the car oving in a circle (usually considered as a Noral force). FNoral Fgravity d. The velocity of the car at position D is deterined using the sae technique as in (b): U i = K +U f gh A = 2 v 2 D + gh D v D = 2g(h A h D ) = 2 9.8(60 40) =9.8 /s e. The force of the track on the car at D can be deterined by using Newton s 2nd Law of Motion to the circular otion: F c = v 2 r F track + F g = v 2 r F track = v 2 F r g Here we re considering down (toward the iddle of the circle) to be in the positive direction, and we re assuing that the force of the track and the force of gravity are both pointing down: the force of gravity is providing soe of the force to keep the car oving in a circle, and the force of the track will provide the reaining. In soe cases, however if the car is traveling very slowly, for exaple a uch saller centripetal force will required, to the point that the track has to provide an upwards force. We would realize that this was the case if we calculated a Force for the track that was negative. Continuing with our calculation: F track = v 2 F r g F track = (200)(9.8)2 20 (200)(9.8) =960N
Practice Test Solutions: f. i. The horizontal braking force causes the velocity of the car to decrease as a function of tie t. The integral describing the car s otion is based on applying Newton s 2nd Law of Motion: F net = a kv = dv dt dv dt = k v dv v = k dt v dv t k = v 0 v dt 0 ii. Now solve the integral for v: v dv t k = v 0 v dt 0 v lnv v0 = k t lnv lnv 0 = k t $ ln v ' & ) = k % v 0 ( t v = e v 0 k t k v = v 0 e t 8. Potential energy can be defined only for conservative forces, and diagras are one way of describing those forces. The relationship between the conservative force acting on a particle, its displaceent, and the potential energy associated with the force is described by the equation ΔU = F x dx, or F x = du dx. a. i. By exaining the graph, the potential energy U is about 5 Joules. ii. Using Conservation of Energy: U i + K i = U f + K f 5J + 0 = 3J + 2 v 2 v = 6 = 6 2 = 2.8 /s iii. At x = 4 there is a position of unstable equilibriu. At this point, based on F x = du, there is dx no acceleration applied by the conservative force (the slope of the U curve is 0), but here, the particle will continue oving past that position, based on the fact that it has 4 Joules of K energy.
Practice Test Solutions: iv. The particle does reach the position x = 7, and because it still has 2 Joules of K energy, it will continue to ove to the right, even though the conservative force is no longer causing any acceleration. b. A particle released fro rest at position x = 8 will not experience any conservative force, and so will not accelerate. It will reain at that location. c. The graph of the conservative force is effectively a graph of the negative slopes of the U-x graph. Key points to ake sure are included in the Force graph are positions where U graph inflects. The Force curve should cross the x-axis at equilibriu points (at x =, 3, and 5 eters, as well as for x > 7 ). Sketch in the rest of the curve based on your interpretation of the U graph.