Transcript File
The elevator in the world’s tallest building in Taiwan will carry a passenger up to the outdoor observations deck on the 89th floor with an acceleration of 0.629 m/s2 . As the elevator goes downwards, the acceleration of the elevator is 0.232m/s2. Philip, who weighs 150. lbs, stands on a bathroom scale in this elevator. 1. What would the scale reading be when the elevator is at rest? 2. What would the scale reading be when he accelerates up to the observation deck on the 89th floor? 3. What would the scale reading be when the elevator is accelerating from the observation deck back to the ground? 4. What would the scale reading be when the elevator is travelling at a constant speed? Warm-Up – (8 points) During the investigation of a traffic accident, police find skid marks along the snowy road that are 125 m long. They determine the coefficient of friction between the car’s tires and the roadway to be 0.28 for the prevailing conditions. Estimate the speed of the car when the brakes were applied. If the speed limit was 55mph, should the driver be ticketed for speeding, ticketed for careless driving (>20 mph over), or not ticketed at all because the speed was safe for the snowy road conditions? Bellwork: In one situation, the ball falls off the top of the platform to the floor. In the other situation, the ball rolls from the top of the platform along the staircase-like pathway to the floor. Indicate what type of forces are doing work upon the ball and whether the energy of the ball is conserved and explain why. Finally, fill in the blanks for the 2-kg ball. Bellwork Solution • The only force doing work is gravity. Since it is an internal or conservative force, the total mechanical energy is conserved. Thus, the 100 J of original mechanical energy is present at each position. So the KE for A is 50 J. • The PE at the same stairstep is 50 J (C) and thus the KE is also 50 J (D). • The PE at zero height is 0 J (F and I). And so the kinetic energy at the bottom of the hill is 100 J (G and J). • Using the equation KE = 0.5*m*v2, the velocity can be determined to be 7.07 m/s for B and E and 10 m/s for H and K. Work Done by a Variable Force Potential Energy, Spring Force and Hooke’s Law Objectives: 1. Define and understand potential energy 2. Calculate the work done by a spring force. 3. Study the WorkEnergy Theorem and apply it in solving problems. Potential Energy: Energy that is stored in an object as a result of its position or condition. • Chemical – energy stored in bonds between atoms • Gravitational – energy stored as a result of an object’s vertical position, or height • Elastic – energy stored in material as a result of stretching or compression Gravitational Potential Energy: depends on the mass and the height of the object. • Potential energy = mass x gravity x height PEgrav m g h • zero height position must first be assigned. • Typically, the ground is considered to be a position of zero height. Example 1: A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of the top of a chair. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? Solution to Example 1 • Given: – m = 3 kg – g = 9.8 m/s2 – h = 0.45 m • PE = m*g*h • PE = (3 kg ) * (9.8 m/s/s) * (0.45 m) • PE = 13.2 J Example 2: If a force of 14.7 N is used to drag the loaded cart (from previous question) along the incline for a distance of 0.90 meters, then how much work is done on the loaded cart? Solution to Example 2 Given: F= 14.7 N d= 0.9 m q = 0o • W = F * d * cos Theta • W = 14.7 N * 0.9 m * cos (0 degrees) • W = 13.2 J Elastic Potential Energy: Spring Forces • Springs are a device which can store elastic potential energy due to either compression or stretching. • A force is required to compress a spring; the more compression there is, the more force which is required to compress it further. • For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k). Calculating the Spring Force Hooke's Law Fspring = k * x where k is the spring constant and x is the amount of stretch or compression. • If a spring is not stretched or compressed, then there is no elastic potential energy stored in it. • The spring is said to be at its equilibrium position. The equilibrium position is the position that the spring naturally assumes when there is no force applied to it. Calculating Elastic Potential Energy The amount of elastic potential energy is related to the amount of stretch (or compression) and the spring constant. The equation is The work done by an external force in stretching or compressing a spring (to overcome the spring force) is calculated by: 1 2 W kx 2 Where k is the spring constant and x is the stretch or compression distance from the equilibrium position Example 3: A spring of spring constant 20 N/m is to be compressed by 0.10 m. • A) what is the maximum force required? • B) What is the work required? Solution to Example 3 • Given: – k = 20 N/m – x = -0.10 m • For a compression, F = -kx • W = ½ kx2 A) F = -(20 N/m)(-0.10 m) = 2.0 N B) W = ½ (20 N/m)(-0.10 m)2 = 0.10 J Round the Clock Spring Problems • Draw a circle, or “clock” on your paper and divide it into 4ths. • Set 4 appointments with your classmates; one at 12 o’clock, one at 3 o’clock, one at 6 o’clock, and one at 9 o’clock. • Solve the following problems at your “appointment”. Round the Clock Spring Problems 3 o’clock: When a 13.2-kg mass is placed on top of a vertical spring, the spring compresses 5.93 cm. Find the force constant of the spring. 6 o’clock: If a spring has a spring constant of 400 N/m, how much work is required to compress the spring 25.0 cm from its undisturbed position? 9 o’clock: A compressed spring that obeys Hooke's law has a potential energy of 18 J. If the spring constant of the spring is 400 N/m, find the distance by which the spring is compressed. 12 o’clock: To compress spring 1 by 0.20 m takes 150 J of work. Stretching spring 2 by 0.30 m requires 210 J of work. Which spring is stiffer? Kinetic Energy: energy of motion • An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. – Vibrational : the energy due to vibrational motion – Rotational: the energy due to rotational motion – Translational: the energy due to motion from one location to another • The amount of translational kinetic energy an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. • The following equation is used to represent the kinetic energy (KE) of an object. 3 o’clock: Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s. 6 o’clock: If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kinetic energy? 9 o’clock: Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed? 12 o’clock: A 900-kg compact car moving at 60 mi/hr has approximately 320 000 Joules of kinetic energy. Estimate its new kinetic energy if it is moving at 30 mi/hr. The Work-Energy Theorem • The net work done on an object is equal to its change in kinetic energy. W = K – K0 = DK • Work is a measure of energy transfer. • Energy is the capacity to do work. • The net force acting on an object causes the object to accelerate, changing its velocity: Combining Kinematic Equations and Newton’s Second Law W F d ma d v W m d 2 x 1 2 1 2 W mv mv0 2 2 2 2 v0 Example 4: The opposing kinetic friction force between a 60.0 kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25.0 m/s, what distance will it slide before coming to a stop? Fnorm Ffric Fgrav Solution to Example 4 • Given: – m = 60.0 kg – v0 = 25.0 m/s Ffric = 50.0 N v = 0 m/s W = K-K0 W = (Fcosq)d = (Ffriccos180)d W = K-K0= ½ mv2 – ½ mv02 (50.0 N)(cos180)(d ) = ½(60.0kg)(0m/s)2-1/2(60.0 kg)(25.0 m/s)2 d = 3.75 x 102 m Law of Conservation of Energy: First Law of Thermodynamics • If the working forces (the forces doing nonzero work) in a system are conservative, the total mechanical energy of the system is conserved. • Conservative Forces: independent of the path but dependent on only the initial and final locations. • Nonconservative Forces: the work done by or against it depends on the path (friction, for example) E = KE + PE KE0+PE0 = KE + PE ½ mv2 + mgh = ½ mv02 + mgh0 Independent Practice • Conservation of Energy – Work with your Best Partner to solve the problems related to total Mechanical Energy. E = KE + PE • Homework: “Work Done by a Variable Force”