Module 4 Work and Energy Part 2 PDF

Positive Work by weight Work done on a body by y1 Gravitational Force (Weight) w y2 Block falling from a height y Negative Work by weight...

Positive Work by weight Work done on a body by y1 Gravitational Force (Weight) w y2 Block falling from a height y Negative Work by weight y2 w Positive Work → Weight same direction y1 with displacement Raising a block to Negative Work → Weight opposite a height y direction with displacement  Gravitational Potential Energy (UG) gravitational energy associated with an object’s position relative to a certain reference point UG = mgy m = mass g = acceleration due to gravity y = vertical position (with respect to a certain reference point)  Gravitational Potential Energy UG1=(1)(9.8)(3) With respect to reference 1 PEG2= (1)(9.8)(5) With respect to Ref 1 reference 2 Ref 2  Conservative Forces (e.g. force of gravity) A force is conservative if the work it does on an object is dependent on the final and initial positions and independent of path taken. Ui Same Uf work !!! Higher work done in moving the ball from the initial position to the final position? Example #12 final initial  Example #12 Answer Same work Same initial and final elevations = same work final initial  Conservative Forces W = - ΔUG = ΔKE UG1 – UG2 = KE2 – KE1 UG1 + KE1 = UG2 + KE2 UG + KE → constant MECHANICAL ENERGY = UG + KE Conservative Forces offer a two-way conversion between kinetic and potential energy  CONSERVATION OF MECHANICAL ENERGY If frictional force and other dissipative mechanisms are disregarded then the total mechanical energy is constant Applicable if the forces in the system are all conservative system ME1 = ME2 UG1 + KE1 = UG2 + KE2  CONSERVATION OF MECHANICAL ENERGY V = 0 m/s U + KE = ME 10 J + 0 J = 10 J PE is 8 J + 2 J = 10 J transformed to KE 2 J + 8 J = 10 J 0 J + 10 J = 10 J  Example #13 A 2-kg ball accidentally fell from the top of building 50 m high. Considering the effect of air resistance is negligible. Calculate the velocity of the ball when he is 10m from the ground. 10  Example #13 UG1 + KE1 = UG2 + KE2 2 mgh1 + 0 = mgh2 + mv /2  Frictionless Roller Coaster Most roller coasters have no engine The conversion of potential energy to kinetic energy is what drives the roller coaster Conservative System: Frictionless Roller Coaster U + KE = ME 10 J + 0 J = 10 J 8 J + 2 J = 10 J 2 J + 8 J = 10 J 0 J + 10 J = 10 J Compute for V (at H=10)? Example #14 Vi=0 v=? H= 50m H= 10m PE=0 Compute for V (at H=10)? Example #13 Vi=0 UG1 + KE1 = ME UG2 + KE2 = ME H= 50m v=? H= 10m PE=0 Solution UG1 + KE1 = UG2 + KE2 2 mgh1 + 0 = mgh2 + mv /2  Frictionless Roller Coaster “D” should be less than “h” D = 2R Conservative System: Pendulum KE = 0 J KE = 0 J UG = 0 Test your understanding! A bowling ball is suspended from the ceiling of a lecture hall by a strong cord. The ball is drawn away from its equilibrium position and released from rest at the tip of the demonstrator’s nose as in the figure. If the demonstrator remains stationary, explain why she is not struck by the ball on its return swing. Would this demonstrator be safe if the ball were given a push from its starting position at her nose? 19 A pendulum consists of a sphere Example #14 of mass m attached to a light cord of length L, as shown in the figure. The sphere is released from rest at point A when the cord makes an angle θA with the vertical, and the pivot at P is frictionless. Find the speed of the sphere when it is at the lowest point B. 20 Example #14 Elastic Potential Energy Work done by a Spring Elastic Potential Energy 23  CONSERVATION OF MECHANICAL ENERGY ME1 = ME2 UG1 + Uel1 + KE1 = UG2 + Uel1 + KE2  Example #15 A block having a mass of 0.80 kg is given an initial velocity vA=1.2 m/s to the right and collides with a spring of negligible mass and force constant k= 50 N/m. Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision.  Example #15 Uel=0 KE=0 a b Spring-loaded popgun Example #16 The launching mechanism of a toy gun consists of a spring of unknown spring constant. When the spring is compressed 0.120 m, the gun, when fired vertically, is able to launch a 0.035 kg projectile to a maximum height of 20.0 m above the position of the projectile before firing. Neglecting all resistive forces, determine the spring constant b KE=0 yb a KE=0 UG=0  The Law of Conservation of Energy E1T = E2T ME1 + Wothers 1 = ME2 + Wothers 2 May include non-conservative forces → Wothers  Work done by Nonconservative Forces Work done depends on the path. No two-way conversion between Potential and Kinetic Energy Mechanical Energy is not conserved A very good example of a non-conservative force is kinetic friction. Crate sliding down a ramp Example #17 A 3.00-kg crate slides down a ramp. The ramp is 1.00m in length and inclined at an angle of 30.0°. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Determine the speed at the bottom of the ramp. MEA + Wothers = MEB A B  Power Time rate of doing work Unit: Watt (W) Power / Average Power Instantaneous Power  Power where F|| is parallel to the displacement Δs 35 A person pulls a block with Example #18 a force of 25N (a) along the horizontal (b) an angle of 0 60 with the horizontal. If the block is moved 6m in the horizontal direction in 4s, how much power is expended? 36 (a) along the horizontal 37  0 (b) At 60 with the horizontal 38 As part of a charity fund raising drive, a runner with Example # 19 mass 50.0 kg runs up the A “power climb” stairs to the top of a 443-m tall building. To lift herself to the top in 15.0 minutes, what must be her average power? 39 Power delivered by an elevator motor Example # 20 An elevator car has a mass of 1 600 kg and is carrying passengers having a combined mass of 200 kg. A constant friction force of 4 000 N retards its motion upward. a) What power delivered by the motor is required to lift the elevator car at a constant speed of 3.00 m/s? b) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.00 m/s2? 40  Force and Potential Energy Force is the negative slope of the Potential Energy – Position graph  Gravitational Potential Energy – Position Graph Slope → Gravitational force / Weight (negative)  Elastic Potential Energy – Position Graph U Hooke’s Law force Fx > 0 Fx < 0 (Elastic force) → towards equilibrium Fx = 0 x Elastic Potential Energy – Position Graph Energy Diagrams Stable equilibrium any minimum in a potential-energy curve Unstable equilibrium any maximum in a potential-energy curve  Example # 20 A particle moves along a line where the potential energy of its system depends on its position r as graphed in the figure. In the limit as r increases without bound, U(r) approaches +1 J. (a) Identify each equilibrium position for this particle. Indicate whether each is a point of stable, unstable, or neutral equilibrium. (b) The particle will be bound if the total energy of the system is in what range? Now suppose that the system has energy -3 J. Determine (c) the range of positions where the particle can be found, (d) its maximum kinetic energy, (e) the location where it has maximum kinetic energy, and (f) the binding energy of the system—that is, the additional energy that it would have to be given in order for the particle to move out to r 🡺∞. A particle moves along a line where the potential energy of its system Example # 20a depends on its position r as graphed in the figure. In the limit as r increases without bound, U(r) approaches +1 J. (a) Identify each equilibrium position for this particle. Indicate whether each is a point of stable, unstable, or neutral equilibrium. Stable equilibrium: Answer r = 1.5 mm and 3.2 mm Unstable equilibrium: r = 2.3 mm Neutral equilibrium: r 🡺∞ (b) The particle will be bound if Example # 20b the total energy of the system is in what range? Answer -5.8 J < U < +1 J Suppose that the system has Example # 20c energy -3 J. (c ) Determine the range of positions where the particle can be found, Answer 0.6 mm < r < 3.6 mm Suppose that the system has Example # 20d energy -3 J. (d) its maximum kinetic energy Answer E = - 3J Umin = - 5.6 J KEmax = E- Umin KEmax = -3 J – (-5.6 J) Umin, KEmax KEmax = 2.6 J (e) the location where it has Example # 20e maximum kinetic energy Answer r = 1.5 mm Suppose that the system has energy Example # 20f -3 J. (f) the binding energy of the system—that is, the additional energy that it would have to be given in order for the particle to move out to r 🡺∞. Answer -3 J + W = 1 J W=4J