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AP Physics C Conservation of Energy Problems Name _____________________________ 5. A 5 kg block is set into motion up an inclined plane with an initial speed of 8 m/s. The block comes to rest after traveling 3 meters up the ramp, which has an angle θ=300. 3m θ a. Determine the net work done on the block Wnet = -160 J b. Calculate the block’s change in potential energy ΔU = +73.5 J c. Calculate the average frictional force exerted on the block. favg = 28.8 N d. What is the coefficient of kinetic friction between the incline and the block? µμk = 0.68 6. A single conservative force F=2x + 4 acts on a 5 kg particle, where x is in meters and F is in Newtons. As the particle moves along the x-axis from x=1 m to x=5 m, calculate each of the following: a. The work done by this force W = 40 J b. The change in potential energy of the particle ΔU = -40 J c. The speed of the particle at x=5 meters if its speed at x=1 m is 3 m/s. v(5) = 3 m/s AP Physics C Conservation of Energy Problems Name _____________________________ 1995M2. A particle of mass m moves in a conservative force field described by the potential energy function U(r) = a(r/b + b/r), where a and b are positive constants and r is the distance from the origin. The graph of U(r) has the following shape. a. In terms of the constants a and b, determine the following. i. The position ro at which the potential energy is a minimum r0 = b ii. The minimum potential energy Uo Umin = 2a b. Sketch the net force on the particle as a function of r on the graph below, considering a force directed away from the origin to be positive, and a force directed toward the origin to be negative. Hint #1: F is the inverse slope of the U vs r curve F = -dU/dr Hint #2: Where is the xintercept of this graph? AP Physics C Conservation of Energy Problems Name _____________________________ The particle is released from rest at r = ro/2 c. In terms of Uo and m, determine the speed of the particle when it is at r = ro . v= d. U0 2m Write the equation or equations that could be used to determine where, if ever, the particle will again come to rest. It is not necessary to solve for this position. Instantaneously: When it has a potential energy equivalent to that at r0/2. Full Rest: Never! e. Briefly and qualitatively describe the motion of the particle over a long period of time. …will oscillate between _____ and _____... Work, Energy, & Power Free Response Question 1 (15 pts) 1. A block of mass m is attached to an ideal spring of spring constant k, the other end of which is fixed. The block is on a level, frictionless surface as shown in the diagram. At time t0, the block is set into simple harmonic motion of period T by an external force pushing it to the right, giving the block initial velocity v0. Express all answers in terms of the given quantities and fundamental constants. v0 m A. Determine the amplitude of the block’s motion. (4 points max) Kinitial = Us,max 1 2 mv0 = 2 1 1 point For applying conservation of energy 1 point For equating the initial kinetic energy of the block with the maximum elastic potential energy 2 kxmax 2 1 point For correctly substituting the definitions of kinetic and elastic potential energy 2 xmax = mv0 k 1 point For the correct answer in terms of the given quantities and fundamental constants B. On the axis below, plot the kinetic energy of the block as a function of time for two periods. Label the vertical axis appropriately. ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power K (J) t (s) T T 2 (4 points max) 2T 3T 2 1 point for drawing any sinusoidal graph sinusoidal graph, positive, with max KE at 0, T/2, T, 3T/2, and 2T 1 point for a graph that never shows negative kinetic energy (graph is always positive) 1 point for showing the maximum kinetic energy is 1 2 mv0 and the 2 minimum kinetic energy is zero 1 point for correctly showing that the maximum kinetic energy occurs at t = 0, T/2, T, 3T/2, and 2T and that the minimum kinetic energy occurs at t = T/4, 3T/4, 5T/4, and 7T/4. ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power C. The block is stopped and a second identical block is glued on top of the first. The blocks are returned to simple harmonic motion with the same initial velocity as before. i. How does the amplitude of the motion of the two blocks together compare to the amplitude found in part a)? Greater than __________ Less than ________ Equal ________ ii. Justify your answer. (3 points max) 1 point For correctly indicating the amplitude increases when the mass increases Greater than __x__ the increased mass increases the kinetic energy when the block is set in motion, meaning the spring will be compressed and stretch more 1 point for any reasonable justification using the equation from part A, as the mass increases so does the amplitude 2 xmax = mv0 k ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power Question 2 (20 pts) A roller coaster car of mass 650 kg is moving with a velocity of 15.0 m/s at the top of the first hill of a roller coaster track as shown in the diagram. The cart rolls without friction down the hill and through a vertical circular loop of radius 12 m. 15.0 m/s 55 m 12 m 30 m A. Calculate the maximum velocity of the roller coaster. 1 point for a correct application of conservation of energy (4 points max) 1 2 mv0 + mgh0 = 2 vmax = 2 2 2 mvmax ⎡ 1 v 2 + gh ⎤ 0 ⎢⎣ 2 0 ⎥⎦ ( ⎡ 1 15 m s ⎢⎣ 2 m = 36 s vmax = vmax 2 1 ) + (9.8 m s ) ( 55m )⎤⎥⎦ 2 2 1 point for recognizing that the maximum kinetic energy occurs when the roller coaster is at the bottom of the first hill (zero potential energy) 1 point for correctly adding the initial kinetic and potential energy to find the total energy 1 point for the correct answer including correct units and reasonable number of significant digits ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power The roller coaster travels without friction through the circular loop. i. On the diagram below, draw and label all of the forces acting on the roller coaster when it is upside down at the top of the loop. 1 point for a correct force vector for the Normal force (3 points max) 1 point for a correct force vector for the weight force FN 1 point for no extraneous vectors Fg ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power ii. Calculate the magnitude of the normal force exerted on the roller coaster when it is upside down at the top of the loop. (6 points max) ∑F = F + Fg = Fcp N FN = mv 1 point for recognizing that the sum of the forces provide a centripetal force on the car 1 point for recognizing that the sum of the forces is the normal force minus the weight force 2 − mg r K top = Etotal − U g ,top ( K = 423, 475 J − ( 650kg ) 9.8 m s 2 1 point for correctly solving for the normal force and substituting the expression for the centripetal force ) ( 30m ) K = 232, 375 J K= v= 1 2 mv 2K 1 point for using conservation of energy to find the velocity 2 2 ( 232, 375 J ) = 650kg m v = 26.74 m FN = FN = mv 1 point for the correctly solving for the velocity s 2 − mg r ( 650kg ) ( 26.74 m s ) 12 m 2 ( − ( 650kg ) 9.8 m s 2 ) 1 point for the correct answer including correct units and reasonable number of significant digits FN = 32, 000 N iii. The safety engineer determines that the acceleration of the riders is too great while they are passing through the loop. Describe a way the engineer can modify the ride to safely reduce the acceleration of the passengers as they go through the loop. Justify your answer. ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power (2 points max) increase the radius, increase height of loop, decrease initial velocity, introduce drag or friction 1 point for any reasonable modification that would reduce the acceleration of the riders through the loop 1 point for an appropriate justification increasing the height of the loop would require the roller coaster to gain more potential energy and thus it would have less kinetic energy and would be traveling slower C. At the end of the ride, the roller coaster car returns to its initial height of 55 m before being brought to a stop by friction. Determine the amount of work that must be done in stopping the roller coaster at this height. (5 points max) Etop = Eend + W f 1 point for any indication of conservation of energy W = ΔK K= 1 2 mv = 2 1 2 ( 650kg ) ( 15 m s ) 1 point for indicating that the work done against friction is equal to the change in energy as the roller coaster comes to a stop 2 K = 73,125 J ΔK = W = −73,125 J 1 point for indicating that at the beginning, the roller coaster has both potential and kinetic energy Alternate solution U g = mgh ( U g = ( 650kg ) 9.8 m s 2 ) (55m ) 1 point for indicating that at the end, the roller coaster has only potential energy U g = 350, 350 J K = 423, 475 J − 350, 250 J 1 point for the correct answer including correct units and reasonable number of significant digits K = 73,125 J ΔK = W = −73,125 J ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power Question 3 (15 pts) An 80 kg object is to be pulled to the top of a 6.0 m ramp by a rope, the other end of which is pulled up by a 2400 W electric winch. The ramp forms a 30° angle with the horizontal. The coefficient of kinetic friction between the ramp and the object is 0.6 and the coefficient of static friction is 0.8. The object moves up the ramp at a constant velocity. 6m 30 ° A. On the diagram of the object showing all the forces on the object. Label each one. (4 points max) below, draw vectors 1 point for a correct force vector for the Normal force FN FT 1 point for a correct force vector for the tension Ff 1 point for a correct force vector for the frictional force Fg 1 point for a correct force vector for the weight force 1 point deducted for any extraneous vectors with a maximum of four points. ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power B. Determine the maximum constant speed at which the winch can pull the object up the ramp. 1 point for any statement indicating that the net force is zero since the speed is constant (7 points max) ∑F x = Fg , x + Ff ,k − FT = 0 FT = Fg , x + Ff ,k 1 point for correctly summing the forces to find the tension FT = mg sin θ + μ k mg cos θ ( FT = ( 80kg ) 9.8 m ( +.6 ( 80kg ) 9.8 m s s 2 2 1 point for correctly substituting expressions for the x-component of weight and the force of friction ) sin ( 30°) ) cos ( 30° ) 1 point for using the coefficient of kinetic friction and not static friction FT = 799 N 1 point for a statement that the maximum constant speed at which the winch can pull the object up the ramp is power or the rate at which work is done P = Fv P = Fv v= P = F v = 3.0 2400 J 1 point for a correct application of P = Fv to find the speed s 799 N m s 1 point for the correct answer including correct units and reasonable number of significant digits ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Work, Energy, & Power C. Determine the total amount of work that the winch must do in pulling the object up the ramp. (4 points max) 1 point for an indication that the net work is the net force times the distance ∑W = ∑ F ⋅ d d = 6m sin ( 30° ) 1 point for indicating the distance the block travels along the ramp = 12 m 1 point for using the net force from part B ∑ W = ( 799 N )(12m ) 1 point for the correct answer including correct units and reasonable number of significant digits W = 9790 J ® Copyright © 2008 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org W=Fcosφ×d 1 f =µFN K= mv² 2 Ug=mgh Ki+ΣW=Kf 1 1 1 2 2 2 Ws = kxi2– kxf 2 Us = kx2 ∆K+∆U =0 Elastic Potential Energy see p.194 The work done by a spring force (Fs = - kx) on a block connected to the spring is given 𝟏 𝟐 𝟏 𝟐 Ws = kxi2 – kxf 2 𝟏 𝟐 Since the elastic potential energy is expressed as Us = kx2 where x=xf -xi Ws = – ∆Us Note that the spring force is a conservative force, so mechanical energy is conserved Q11) A 1.6kg block on a frictionless surface is pressed against a spring with a spring constant of k=1000N/m. The pressure causes the spring to compress 0.02m and then released. What is the kinetic energy of the moment ball passes the equilibrium position? What is the speed at that position? see solution below for your reference ii) Solution using Conservation of Mech. E method i) Solution using W-E method Ki + ΣW = Kf Only force doing work is the spring force Fs => Ki + Ws = Kf Ki + Usi = Kf + Usf , here Ki= Usf =0 => Usi = Kf 1 => 2 𝑘𝑥𝑖2 = Kf If we choose the initial state when the moment the block is 1 1 released, then Ki =0, Ws=2 𝑘𝑥𝑖2 − 2 𝑘𝑥𝑓2 and xi=0.02m, xf=0 Ans)0.2J, 0.5m/s 1 => 2 𝑘𝑥𝑖2 = Kf then find Kf and solve for vf 7 W=Fcosφ×d f =µFN 1 K= mv² 2 Ug=mgh Ki+ΣW=Kf 1 1 1 2 2 2 Ws = kxi2– kxf 2 Us = kx2 ∆K+∆U =0 Q12) The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6m/s. How fast is the block moving at the instant the spring has been compressed 0.15m? k=2000N/m, mass is 2kg. xf =0.15m If we choose the initial state when the moment block comes in contact with spring, then xi=0 and xf=0.15m. vf =? xi=0 vi =6m/s Ans) v=3.67m/s Q13) revisit! A crate of mass 12kg slides from rest down a frictionless 35° incline and is stopped by a strong spring with k=3×104N/m. The block slides 3m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed? 3m Ans) 0.116m 8 Conservative and Nonconservative Forces (see p.196 for more details) Conservative Forces 1. A force is conservative if the work it does on a particle moving between any two points is independent of the path taken by the particle ex) Fg is a conservative force because it is path independent 2. The work done by a conservative force on a particle moving through any closed path is zero. ( A closed path is one in which the beginning and end points are identical) - Examples of conservative forces are gravitational force (Fg) and elastic force(Fs ). - For an object free-falling from a certain height, the total mechanical energy(E) of the system consists of kinetic and gravitational potential energy. - The total mechanical energy(E) of a falling object remains constant in any isolated system if the force acting on the object is a conservative force - If an object is free-falling, then Ki + Wg = Kf Since the force of gravity is conservative, Wg = ‒∆Ug => ∆K + ∆U = 0 for conservative force (∆Etot = 0) => E = Ki + Ugi = Kf + Ugf **Non-conservative Forces - If the forces acting on objects within a system are conservative, then the total mechanical energy of the system remains constant. However, if some of the forces acting on objects within the system are not conservative, then the total mechanical energy does not remain constant - A force is non-conservative if it causes a change in mechanical energy E, which is the sum of kinetic and potential energies. Examples of non-conservative forces are an applied force outside the isolated system and frictional force i) Work Done by an Applied Force ( = an external force exerted outside the system) - If you lift a book over a certain distance(Wapp), then Ki + Wapp + Wg = Kf Since, work done by gravity(Wg) is conservative, Wg= - ∆Ug, so Ki + Wapp ‒ ∆Ug = Kf => Wapp = ∆K + ∆Ug and Wapp = ∆Etot for non-conserative force (∆Etot ≠ 0) - We can see that the total mechanical energy is not conserved because the Wapp changes ∆Etot. 9 ii) Work Done by Kinetic Friction - If an object moves a distance d on a flat surface, the only force that does work is the force of kinetic friction Ki + Wfriction = Kf Ki – fk d = Kf => ∆K = - fkd - If the book moves on an incline that has friction, a change in the gravitational potential energy of the book-Earth system also occurs ∆E = ∆K + ∆Ug = - fkd - Since frictional force causes a change in mechanical energy, it is non-conservative force. Summary W=Fcosφ×d f =µFN 1 K=2mv² 1 Ug=mgh Us=2kx² 1 Ki+ΣW=Kf 1 Ws = 2kxi2– 2kxf 2 If there is no friction or an applied force outside the system and only conservative force exists => ∆K + ∆U = 0 If there is friction acting in the system =>∆K + ∆U = - fkd If there is an applied force from outside the system on a frictionless surface => ∆K + ∆U = Wapp If there is an applied force from outside the system on a rough surface => ∆K + ∆U = Wapp ‒ fkd FN Q14) compare with Q4! A block of mass 2kg is sliding down a rough inclined surface. The block begins at rest 2m above the plane surface. The friction acting on the block is 4N. The angle of the incline is 30°. Find the speed when the block reaches the bottom using Conservation of Energy Theorem. That is, use ∆E = ∆K + ∆Ug = - fkd. (solution is on the next page) 10 fk φ h=2m 30º Fg Solution) ∆K + ∆Ug = - fkd. =>Kf - Ki + Ugf - Ugi = - fkd , here Ki=Ugf =0 => Kf - Ugi = - fkd So rearrange according to Kf 1 => 𝑚𝑣𝑓2 =mghi - fkd , then solve for vf ans) 4.82m/s 2 W=Fcosφ×d 1 f =µFN K=2mv² Ug=mgh ∆K + ∆U =0: for conservative forces Ki+ΣW=Kf 1 1 Ws = 2kxi2– 2kxf 2 1 Us = 2kx2 ∆K + ∆U = Wapp ‒ fkd: for non-conservative forces Q15) compare with Q11! A 1.6kg block on a rough surface is pressed against a spring with a spring constant of k =1000N/m. The pressure causes the spring to compress 0.02m and then released. A constant frictional force of 4N retards its motion from the moment it is released What is the speed the moment the block passes equilibrium position? Ans) 0.387m/s Q16) A 10kg block on a horizontal frictionless surface is attached to a light spring(k=200N/m). The block is initially at rest at its equilibrium position when a horizontal force Fp=80N pulls on the block. What is the speed of the block when it is 0.13m from its equilibrium position? Fp Ans) 1.32m/s 11 W=Fcosφ×d 1 f =µFN K=2mv² Ug=mgh ∆K + ∆U =0: for conservative forces Ki+ΣW=Kf 1 1 Ws= 2kxi2– 2kxf 2 1 Us = 2kx2 ∆K + ∆U = Wapp - fkd: for non-conservative forces m1=3kg Q17) Two boxes are connected to each other on a frictionless surface as shown. The system is released from rest and the m2=1kg box falls through a vertical distance of h=1m. What is the kinetic energy of the m2 just before it reaches the floor? Solve using i) Newton’s 2nd law, ii) Work.Kinetic Energy method and iii) Conservation of Energy method m2=1kg 1m Solution is on the next page for your reference 12 W=Fcosφ×d 1 f =µFN K=2mv² Ki+ΣW=Kf Ug=mgh ∆K + ∆U =0: for conservative forces 1 1 Ws= 2kxi2– 2kxf 2 1 Us = 2kx2 ∆K + ∆U = Wapp - fkd: for non-conservative forces FN Method I. Using Newton's Law m1 : ∑Fx= T = m1a ∑Fy= FN - m1g = 0 m2 : ∑Fy= m2g – T= m2a m1=3kg T m1g T m2=1kg From the above two equations, eliminating tension force 'T' and solving for acceleration 'a' gives 1m 𝑚2 𝑔 1 +𝑚2 a=𝑚 m2g = 2.45m/s2 Using 2ad = vf 2 – vi 2 to find vf, where vi=0 and replacing 'd' for 'h' gives vf =√2𝑎ℎ = 2.21m/s Since K= 1 mv2, 2 Kf = 1 ×1×2.212=2.44J 2 m1=3kg T Method II. Using Work.Kinetic Energy. Set up the equations for m1 and m2 separately where the tension force T is an applied force from outside the system. T Isolated System for m1! m2=1kg Isolated system for m2! m1: Ki+WT =Kf =>0+Thcos0°=½m1v1f 2 =>Th=½mv1f 2 -----------------(1) 1m m2g m2: Ki+WT +Wg=Kf =>0+ Thcos180°+Fghcos0°=½m2v2f 2 => – Th+mgh=½m2v2f 2 -----------------(2) The speed of m1 and m2 are the same since they are connected with the string, so setting v1=v2=v, combining (1) and (2)gives ½(m1+m2)vf 2=m2gh Solving for vf gives vf =2.21m/s 1 2 1 2 Since K= mv2, Kf = ×1×2.212=2.44J 13 AP Physics C 3.2 a Chapter 03 1. George pushes in a spring with spring constant k a distance ∆x from its neutral position. (a) What force does the spring exert on George? (b) What is this rule called? (c) What are the SI units of k? F = −k∆x Hooke’s Law N/m (d) What is the work George has done on the spring? (e) What is the potential energy of the spring? 3.2 a U= W = 1 k∆x2 2 1 k∆x2 2 2. George pulls a spring with spring constant k a distance ∆x from its neutral position. (a) What force does the spring exert on George? (b) What is this rule called? (c) What are the SI units of k? F = −k∆x Hooke’s Law N/m (d) What is the work George has done on the spring? (e) What is the potential energy of the spring? 3.2 a Section 02 U= W = 1 k∆x2 2 1 k∆x2 2 3. George pushes in a spring with spring constant k = 200 N/m a distance ∆x = 5 m from its neutral position. F = −1000 N (a) What force does the spring exert on George? (b) From whence was this result derived? (c) What are the SI units of k? Hooke’s Law N/m (d) What is the work George has done on the spring? (e) What is the potential energy of the spring? 3.2 a W = 2500 J U = 2500 J 4. George pulls a spring with spring constant k = 150 N/m a distance ∆x = 9 from its neutral position. F = −1350 N (a) What force does the spring exert on George? (b) What is this rule called? (c) What are the SI units of k? Hooke’s Law N/m (d) What is the work George has done on the spring? (e) What is the potential energy of the spring? W = 6075 J U = 6075 J 3.2 b 5. A force is conservative if: The work done by the force on an object does NOT depend not the path taken. 3.2 b 6. George lift a object of mass m = 12 kg a hight of h = 12 m off the ground. He then lowers it to the ground. (a) What work has George done on the box? (b) Justify your answer. 3.2 c None Gravity is a conservative force. 7. George pushes a block up an incline to a height h. Starting at the same point, Georgina just lifts her identical block to the same height. Assume the incline is frictionless. Page 2 (a) Which block has had more work done on it? They are the same (mgh). (b) Justify your answer. Gravity is a conservative force. Work does by it not depend on the path taken. 3.2 c 8. George pushes a block up an incline to a height h. Starting at the same point, Georgina just lifts her identical block to the same height. Assume the incline has friction. (a) Which block has had more work done on it? George has done more work. (b) Justify your answer. Although gravity is a conservative force, friction is not. The work done by it depends on the path taken. Page 3 A.P. Physics 1 First Semester Review Sheet, Page 7 Chapter 7: Work and Kinetic Energy A. Work • A force exerted through a distance performs mechanical work. • When force and distance are parallel, W = Fd with Joules (J) or Nm as the unit of work. • When force and distance are at an angle, only the component of force in the direction of motion is used to compute the work: W = ( F cos q )d = Fd cos q • Work is negative if the force opposes the motion ( q >90o). Also, 1 J = 1 Nm = 1 kg-m2/s2. • n If more than one force does the work, then WTotal = å Wi i =1 1 1 WTotal = DK = K f - Ki = mv 2f - mvi2 2 2 • The work-kinetic energy theorem states that • See Table 6 for more information about kinetic energy. • In thermodynamics, æ Aö æ F ö W = Fd = Fd ç ÷ = ç ÷ ( Ad ) = PDV for work done on or by a gas. è Aø è Aø Kinetic Energy Type Table 6: Kinetic Energy Equation Kinetic Energy as a Function of Motion: K= Kinetic Energy as a Function of Temperature: 3 æ1 2ö ç mv ÷ = Kave = kT 2 è2 øave Comments Used to represent kinetic energy in most conservation of mechanical energy problems. 1 2 mv 2 Kinetic theory relates the average kinetic energy of the molecules in a gas to the Kelvin temperature of the gas. B. Determining Work from a Plot of Force Versus Position • In a plot of force versus position, work is equal to the area between the force curve and the displacement on the x-axis. For example, work can be easily computed using W = Fd when rectangles are present in the diagram. • For the case of a spring force, the work to stretch or compress a distance x from equilibrium is 1 2 kx . On a plot of force versus position, work is the area of a triangle with base x 2 (displacement) and height kx (magnitude of force using Hooke’s Law, F = kx ). W= C. Determining Work in a Block and Tackle Lab • The experimental work done against gravity, WLoad , is the same as the theoretical work done by the spring scale, WScale . • WOutput = WLoad = Fd Load = W d Load = mgd Load • WInput = WScale = Fd Scale where F = force read from the spring scale and d Scale = distance the scaled • moved from its original position. Note that the force read from the scale is ½ of the weight when two strings are used for the pulley system, and the force read is ¼ of the weight when four strings are used. where d Load = distance the load is raised. A.P. Physics 1 First Semester Review Sheet, Page 8 D. Power W t • P= • 1 W = 1 J/s P = Fv or and with Watts (W) as the unit of Power. 746 W = 1 hp where hp is the abbreviation for horsepower. Chapter 8: Potential Energy and Conservation of Energy A. Conservative Forces Versus Nonconservative Forces 1. Conservative Forces • A conservative force does zero total work on any closed path. In addition, the work done by a conservative force in going from point A to point B is independent of the path from A to B. In other words, we can use the conservation of mechanical energy principle to solve complex problems because the problems only depend on the initial and final states of the system. • In a conservative system, the total mechanical energy remains constant: Ei = E f . Since E = U + K , it follows that U i + K i = U f + K f . See Table 6 for kinetic energy, K , and Table 7 for potential energy, U , for additional information. • For a ball thrown upwards, describe the shape of the kinetic energy, potential energy, and total energy curves on a plot of energy versus time. • Examples of conservative forces are gravity and springs. 2. Nonconservative Forces • The work done by a nonconservative force on a closed path is not zero. In addition, the work depends on the path going from point A to point B. • In a nonconservative system, the total mechanical energy is not constant. The work done by a nonconservative force is equal to the change in the mechanical energy of a system: WNonconservative = Wnc = DE = E f - Ei . • Examples of nonconservative forces include friction, air resistance, tension in ropes and cables, and forces exerted by muscles and motors. Potential Energy Type Gravitational Potential Energy: Gravitational Potential Energy Between Two Point Masses: Table 7: Potential Energy Equation U = mgh U = -G m1m2 r Good approximation for an object near sea level on the Earth’s surface. where G = 6.67 x 10-11 Nm2/kg2 = Universal Gravitation Constant Elastic Potential Energy: Comments Works well at any altitude or distance between objects in the universe; recall that r is the distance between the centers of the objects. Useful for springs, rubber bands, 1 2 kx where k is the force bungee cords, and other 2 stretchable materials. (spring) constant and x is the U= distance the spring is stretched or compressed from equilibrium.