Potential Energy and Conservation of Energy Chap. 7 & 8

Level : AP Physics Potential Energy and Conservation of Energy Chap. 7 & 8 Potential Energy of a System see p.191 in the textbook - Potential energy is the energy associated with the arrangement of a system of objects that exert forces on each other - Potential energy can be thought of as energy of position - Potential energy can be thought as stored energy that can either do work or be converted to kinetic energy Gravitational Potential Energy & Conservation of Mechanical Energy - Gravitational Potential Energy(Ug) is defined as Ug = mgy - Conservation of Mechanical Energy is described as The total mechanical energy(e) of a system remains constant in any isolated system if the force acting on the object is a conservative force *~conservative force is explained in details on p.9 in the worksheet E tot = K i + U i = K f + U f or ΔK + ΔU =0 where ΔE tot = 0 - An object is falling towards the Earth where the initial height at some instantaneous point is y i and the final height is y f. (air resistance is ignored) y i y=y i -y f - The gravitational force F g = mg does work on the object. => K i + ΣW = K f. (Work.Kinetic Energy Theorem ) => K i +W g = K f. (The only force doing work on the object is force of gravity) - If y= y f - y i, and since the force of gravity(f g ) is doing work on the object over a distance of y, y f W g = mgy i mgy f = U i U f = U g W g = U g : work done by gravity = the negative change in U g => K i +W g = K f => K i U g = K f => K + U g = 0 or E tot = K i + U i = K f + U f 1

If you drop a ball from a certain height, the gravitational potential energy(u g ) decreases and the kinetic energy(k) increases, but the total mechanical energy is conserved. m=10kg, h=10.2m, g=9.8m/s 2 U g =1000J K=0J *~ U g = mgh K = ~* Total Energy = U g + K = 1000J h U g =500J K=500J Total Energy = U g + K = 1000J U g =250J K=750J Total Energy = U g + K = 1000J U g =0J K=1000J Total Energy = U g + K = 1000J W=Fdcosφ f = µf N K = mv² U g = mgh K i +ΣW=K f K i + U i = K f + U f Q1) An object of mass 10kg is lifted at a height of 10m. i) What is the gravitational potential energy of the object at that height? What is the kinetic energy at that height? What is the total mechanical energy? ii) When the object falls 8m, what is the gravitational potential energy and kinetic energy at that height? What is the speed at that height? What is the total mechanical energy? Q2) A 0.04kg body starting from rest falls through a vertical distance of 0.25m to the ground. (a) What is the kinetic energy of the body just before it hits the ground? (b) What is the speed of the body just before it hits the ground? 2

W=Fdcosφ f = µf N K = mv² U g = mgh K i +ΣW=K f K i + U i = K f + U f Q3) A boy throws up a 0.15kg stone from the top of a 20m cliff with a speed of 15m/s. Find its kinetic energy and speed when it lands in a river below. U g,max, K=0 U g + K U g =0, K max Ans) 46.3J, 24.8m/s Q4) A block of mass 2kg is sliding down a frictionless inclined surface. The block begins at rest 2m above the plane surface. The angle of the incline is 30. i) Find the speed when the block reaches the bottom using Work.Kinetic Energy Theorem. h=2m φ F N 30º F g Solution) K i + ΣW = K f Only force doing work on block is gravity F g => K i + W g = K f Since Work is defined as W=Fcosφd => W g =F g cos(90º-30º)d, so K i + F g cos60ºd = K f Since the block starts at rest K i =0 => 0 + mgcos60ºd = Finally, solve for v f (remember that h=dsin30º) ans) 6.26m/s 3

W=Fdcosφ f = µf N K = mv² U g = mgh K i +ΣW=K f K i + U i = K f + U f ii) Find the speed when the block reaches the bottom using Conservation of Mechanical Energy Theorem. Solution) Since gravitational force F g is conservative, mechanical energy is conserved throughout the motion From K i + U gi = K f + U gf, K i =U gf =0, so U gi = K f!! => mgh =,then solve for v f Energy is Path Independent Mechanical Energy is conserved when a conservative force does work on an object in an isolated system. If there is an applied force from outside of the system or frictional force, then mechanical energy is not conserved. A block is released at height of h=5m for all three cases. Using K i +ΣW=K f, find the speed of the block at the bottom for each case. d will be the distance traveled. θ=30 θ=60 case-1 case-2 case-3 Solution) K i + ΣW = K f where K i =0 Only force doing work on block is gravity F g => W g = K f => (mgcosφ)d= since φ=90 θ, cosφ=sinθ and sinθ=h/d => mgh=. So v f = = = 9.9m/s for all cases 4

W=Fdcosφ f = µf N K = mv² U g = mgh K i +ΣW=K f K i + U i = K f + U f Q5) A child rides on a smooth slide of a height of 2m. The child starts from rest at the top. Determine the speed at the bottom. Ans) v=6.3m/s *~If the path is not linear, it is impossible to use W-K theorem because φ is a variable! Q6) A roller coaster starts from rest at the top of an 18m hill. The car travels to the bottom of the hill and continues up the next hill that is 10m high. How fast is the car moving at the top of the 2 nd hill if the friction is ignored? h=18m h=10m Ans) v=12.5m/s Q7) A roller coaster starts from rest at position 1 shown below and rolls without friction along the loop. The roller coaster first pass A, goes around the loop and passes B and then passes C. The height h=1m and the radius of the loop is r=0.4m. Find the speed of the roller coaster at position A, B and C. h B r A C Ans) A&C is 4.43m/s, B is 1.98m/s 5

W=Fdcosφ f = µf N K = mv² U g = mgh K i +ΣW=K f K i + U i = K f + U f Q8) A 0.8kg pendulum bob on a 2m cord is pulled sideways until the cord makes an angle of 36.9 with the vertical. Find the speed of the bob as it passes through the position B after being released at rest. Position A is the maximum height and the position B will the low point of the bob. B 36.9 A h a) 2.8m/s b) 3.6m/s c) 4.5m/s d) 6.1m/s Q9) A skier leaves the top of a slope with an initial speed of 5m/s. Her speed at the bottom of the slope is 13m/s. What is the height of the slope? (*~remember that the direction of the speed is irrelevant when using 'Conservation of Mechanical Energy' equation) v i =5m/s a) 1.14m b) 4.61m c) 6.45m d) 7.35m h=? v f =13m/s Q10) A projectile of mass 2kg is fired with initial speed of 50m/s at an angle of θ=53.1 from the horizontal surface. *~v x =vcosθ, v y =vsinθ~* i) What is the initial kinetic energy? a) 900J b) 1600J c) 2500J d) 0J ii) What is the kinetic energy when the projectile is at the highest point in its trajectory? a) 900J b) 1600J c) 2500J d) 0J (*'0J ' is not the answer!!!) iii) What is the gravitational potential energy at the highest point in its trajectory? a) 900J b) 1600J c) 2500J d) 0J iv) Find the maximum height of the trajectory. a) 81.6m b) 62.9m c) 49.9m d) 35.4m 6

W=Fcosφ d f =µf N K= mv² U g =mgh K i +ΣW=K f W s = kx i 2 kx f 2 U s = kx 2 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 W s = kx i 2 kx f 2 Since the elastic potential energy is expressed as Us = kx 2 where x=x f -x i W s = U s 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 i) Solution using W-E method K i + ΣW = K f Only force doing work is the spring force F s => K i + W s = K f If we choose the initial state when the moment the block is released, then K i =0, W s = and x i =0.02m, x f =0 ii) Solution using Conservation of Mech. E method K i + U si = K f + U sf, here K i = U sf =0 => U si = K f => = K f Ans)0.2J, 0.5m/s => = K f then find K f and solve for v f 7

W=Fcosφ d f =µf N K= mv² U g =mgh K i +ΣW=K f W s = kx i 2 kx f 2 U s = kx 2 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. If we choose the initial state when the moment block comes in contact with spring, then x i =0 and x f =0.15m. x f =0.15m x i =0 v f =? v i =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 10 4 N/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 Conservative Forces (see p.196 for more details) 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) F g 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 (F g ) and elastic force(f s ). - 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 K i + W g = K f Since the force of gravity is conservative, W g = U g => K + U = 0 for conservative force ( E tot = 0) => E = K i + U gi = K f + U gf **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(w app ), then K i + W app + W g = K f Since, work done by gravity(wg) is conservative, Wg= - Ug, so K i + W app U g = K f => W app = K + U g and W app = E tot for non-conserative force ( E tot 0) - We can see that the total mechanical energy is not conserved because the W app changes E tot. 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 K i + W friction = K f K i f k d = K f => K = - f k d - 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 + U g = - f k d - Since frictional force causes a change in mechanical energy, it is non-conservative force. Summary W=Fcosφ d f =µf N K= mv² U g =mgh U s = kx² K i +ΣW=K f W s = kx 2 2 i kx f 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 = - f k d If there is an applied force from outside the system on a frictionless surface => K + U = W app If there is an applied force from outside the system on a rough surface => K + U = W app f k d 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 + U g = - fkd. (solution is on the next page) F N f k φ h=2m 30º F g 10

Solution) K + U g = - f k d. =>K f - K i + U gf - U gi = - f k d, here K i =U gf =0 => K f - U gi = - f k d So rearrange according to K f => =mgh i - f k d, then solve for v f ans) 4.82m/s W=Fcosφ d f =µf N K= mv² U g =mgh K i +ΣW=K f W s = kx 2 2 i kx f U s = kx 2 K + U =0: for conservative forces K + U = W app f k d: 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 F p =80N pulls on the block. What is the speed of the block when it is 0.13m from its equilibrium position? F p Ans) 1.32m/s 11

W=Fcosφ d f =µf N K= mv² U g =mgh K i +ΣW=K f W s = kx 2 2 i kx f U s = kx 2 K + U =0: for conservative forces K + U = W app - f k d: for non-conservative forces Q17) Two boxes are connected to each other on a frictionless surface as shown. The system is released from rest and the m 2 =1kg box falls through a vertical distance of h=1m. What is the kinetic energy of the m 2 just before it reaches the floor? Solve using i) Newton s 2 nd law, ii) Work.Kinetic Energy method and iii) Conservation of Energy method Solution is on the next page for your reference m 1 =3kg 1m m 2 =1kg 12

W=Fcosφ d f =µf N K= mv² U g =mgh K i +ΣW=K f W s = kx 2 2 i kx f U s = kx 2 K + U =0: for conservative forces K + U = W app - f k d: for non-conservative forces Method I. Using Newton's Law m 1 : F x = T = m 1 a F y = F N - m 1 g = 0 m 2 : F y = m 2 g T= m 2 a F N m 1 g m 1 =3kg T T From the above two equations, eliminating tension force 'T' and solving for acceleration 'a' gives 1m m 2 g m 2 =1kg a = = 2.45m/s 2 Using 2ad = v f 2 v i 2 to find v f, where v i =0 and replacing 'd' for 'h' gives Since K= mv 2, K f = 1 2.21 2 =2.44J v f = = 2.21m/s Method II. Using Work.Kinetic Energy. Set up the equations for m 1 and m 2 separately where the tension force T is an applied force from outside the system. m 1 : K i +W T =K f =>0+Thcos0 =½m 1 v 1f 2 =>Th=½mv 1f 2 -----------------(1) m 1 =3kg T Isolated System for m 1! Isolated system for m 2! 1m T m 2 g m 2 =1kg m 2 : K i +W T +W g =K f =>0+ Thcos180 +F g hcos0 =½m 2 v 2f 2 => Th+mgh=½m 2 v 2f 2 -----------------(2) The speed of m 1 and m 2 are the same since they are connected with the string, so setting v 1 =v 2 =v, combining (1) and (2)gives ½(m 1 +m 2 )v f 2 =m 2 gh Solving for v f gives v f =2.21m/s Since K= mv 2, K f = 1 2.21 2 =2.44J 13

How you choose your system makes a big difference The tension force T can be an applied force or an internal force depending on how you choose your system. If you choose each mass separately as an isolated system, the tension force must be solved as an applied force( =similar to method II) and the mechanical energy is not conserved. If you choose the system that encompasses both masses as below, then the tension force becomes an internal force and the mechanical energy is conserved. Method III. Using Energy Conservation (system as whole) First, the gravitational potential energy and kinetic energy for m 1 is 'U g1 ' and 'K 1 '. And the gravitational potential energy and kinetic energy for m 2 is 'U g2 ' and 'K 2 ' Conservation of mechanical energy states single system!! m 1 =3kg 1m m 2 =1kg K 1i +U g1i + K 2i +U g2i = K 1f +U g1f + K 2f +U g2f Since U g1i = U g1f, and K 1i =K 2i =0 U g2i = K 1f + K 2f m 2 gh i = + The speed of m 1 and m 2 are the same since they are connected with the string, so v 1 =v 2 =v, Solving for v f gives Since K= mv 2, K f = 1 2.21 2 =2.44J m 2 gh i = + v f =2.21m/s Relationship between Conservative Forces and Potential Energy When an object is free-falling in the air, the only force acting on the object is F g (assuming air resistance is negligible). Since F g is a conservative force, the work done by F g will be equal to the negative change in gravitational potential energy( U g ). W g = U g : work done by gravity = the negative change in U g This can also be applied to a spring-block system, where there is no friction and applied force acting in the system. W s = U s : work done by spring = the negative change in U s 14

In general, the work done by internal conservative force will be equal to the negative change in potential energy W int = = U where F x is a conservative force. The above relationship between work and force can be rewritten as F x = Q18) To stretch a certain nonlinear spring by an amount x requires a force F given by F x = 40x - 6x 2, where F x is in newtons and x is in meters. What is the change in potential energy when the spring is stretched 2 meters from its equilibrium position? a) 16 J b) 28 J c) 56 J d) 64 J e) 80 J Q19) A single conservative force F x = (6.0x 12) N (x is in m) acts on a particle moving along the x axis. The potential energy associated with this force is assigned a value of +20J at x=0. What is the potential energy at x=3.0m? a) +11J b) +29J c) +9J d) -9J e) +20J Q20) A 0.40kg particle moves under the influence of a single conservative force. At point A where the particle has a speed of 10m/s, the potential energy associated with the conservative force is +40J. As the particle moves from A to B, the force does +25J of work on the particle. What is the value of the potential energy at point B? a) +65J b) +15J c) +35J d) +45J e) -40J 15

Power see p.232 Two cars were climbing a roadway hill and both reached the top, but one car reached the top in lesser time than the other. Both cars have done equal amount of work, but the rate amount of work done is different. From a practical viewpoint, it is interesting to know not only the work done by vehicles but also the rate at which it is done. The time rate of doing work is call power. The average power is defined as Instantaneous power can be expressed as = The standard unit for power is joules per second(j/s), also call watt(w) One horsepower(hp) is equal to 746 watts, so 1W = 1J/s =1kg m 2 /s 3 1hp = 746W Example) A 300W light bulb running for 12 hours would be converted to 0.3kW 12h = 3.6kWh of electrical energy used. Or 300W 12 3600s=1.3 10 7 J of energy used in 12 hours. (Remember that the energy is power used multiplied by the time during which it was used) 1hp = 746W Q1) A 700N Marine in basic training climbs a 10m vertical rope at a constant speed in 8s. What is his power output? a) 875W b) 724W c) 650W d) 572W 16

1hp = 746W Q2) A certain automobile engine delivers 2.24 10 4 W(=30hp) to its wheels when moving at a constant speed of 27m/s. What is the resistive force acting on the automobile? a) 540N b) 664N c) 750N d) 830N Q3) A skier of mass 70kg is pulled up a smooth frictionless slope by a motor-driven cable. i) How much work is required for him to be pulled a distance of 60m up a 30 slope at a constant speed of 2m/s? a) 1.30 10 4 J b) 2.06 10 4 J c) 2.70 10 4 J d) 3.35 10 4 J ii) A motor of what power is required to perform this task? a) 450W b) 569W c) 686W d) 734W Q4) Power Delivered by an Elevator Motor An elevator car has a mass of 1000kg and is carrying passengers having a combined mass of 800kg. A constant frictional force of 4000N retards its motion upward. What must be the minimum power delivered by the motor to lift the elevator car at constant speed of 3m/s? a) 6.48 10 4 W b) 4.23 10 4 W c) 2.78 10 4 W d) 1.06 10 4 W 3m/s 17