Physics 2A Chapter 6: Work and Energy “It is good to have an end to journey toward; but it is the journey that matters, in the end.” Ursula K. Le Guin “Nobody made a greater mistake than he who did nothing because he could only do a little.” Edmund Burke

Reading: pages 142 - 165 (skip section 6.9) Outline: ⇒ work done by a constant force ⇒ kinetic energy ⇒ work-energy theorem ⇒ gravitational potential energy ⇒ conservation of mechanical energy ⇒ conservative and non-conservative forces ⇒ work-energy theorem with non-conservative forces

Problem Solving You should know how to calculate the work done by a force if the force is constant. In some cases, you might need to solve a Newton's second law equation to find the force. The force in question might be the only force acting on an object or one of several. The work-kinetic energy theorem tells us that the net work W done on a particle is equal to the change in the kinetic energy of the particle. That is, W = ∆K, or since the kinetic energy is given by ½ mv2, W = ½ m(v2f - v2o). Here m is the mass of the particle, vo is its speed at the beginning of the interval, and vf is its speed at the end of the interval. Remember that W is the net work done by all forces, or the work done by the net force. In some problems you are asked to find the net work done on an object, given its initial and final speeds (and its mass). This is a direct application of the work-kinetic energy theorem. In other problems you will use the definition of work to calculate its value, given the force and displacement, then use the work-kinetic energy theorem to find the final speed, given the initial speed.

Problems involving the conservation of mechanical energy are relatively straightforward. You must first select an object to analyze. You must then decide if all the forces on the object are conservative. The force of gravity is a conservative force; the force of friction and almost all other forces are non-conservative. If all forces that do work are conservative, then conservation of mechanical energy yields Ko + PEo = Kf + PEf, where Ko and PEo are the kinetic and potential energies at one point in time and Kf and PEf are the kinetic and potential energies for another point in time. You will be given enough information to calculate three of the four energies that appear in the energy equation and can use the conservation of mechanical energy to compute the fourth. If there are nonconservative forces acting on the object, then the appropriate expression is WNC = Ef – Eo or WNC = (KEf + PEf) – (KEo + PEo) In many problems you are asked to find some parameter that occurs in the expression for the kinetic or potential energies. For example, you may be asked for the height of an object, the speed of an object, or its mass. You first find the relevant energy (kinetic or potential), then solve for the parameter.

Questions and Example Problems from Chapter 6 Question 1 A net external force acts on a particle. This net force is not zero. Is this sufficient information to conclude that (a) the velocity of the particle changes, (b) the kinetic energy of the particle changes, and (c) the speed of the particle changes? Give your reasoning in each case.

Question 2 In the figure below, a block slides from A to C along a frictionless ramp, and then it passes through horizontal region CD, where a frictional force acts on it. Is the block's kinetic energy increasing, decreasing, or constant in (a) region AB, (b) region BC, and (c) region CD? (d) Is the block's mechanical energy increasing, decreasing, or constant in those regions?

Problem 1 To pull a 50 kg crate across a horizontal floor at a constant velocity, a worker applies a force directed 20o above the horizontal. A 25.0 N frictional force opposes the motion of the crate. As the crate moves 3.0 m, what is the work done on the crate by (a) the worker’s force, (b) the kinetic frictional force, (c) the gravitational force on the crate, and (d) the normal force? (e) What is the total work done on the crate?

Problem 2 A 0.075 kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow?

Problem 3 A rescue helicopter lifts a 79-kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance of 11 m. (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person’s weight? (d) Use the work-energy theorem and find the final speed of the person.

Problem 4 A shot-putter puts a shot (weight = 71.1 N) that leaves his hand at a distance of 1.52 m above the ground. (a) Find the work done by the gravitational force when the shot has risen to a height of 2.13 m above the ground. (b) Determine the change in the gravitational energy of the shot.

Problem 5 A slingshot fires a pebble from the top of a building at a speed of 14.0 m/s. The building is 31.0 m tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

Problem 6 A water-skier lets go of the tow rope upon leaving the end of a jump ramp at a speed of 14.0 m/s. As the drawing indicates, the skier has a speed of 13.0 m/s at the highest point of the jump. Ignoring air resistance, determine the skier’s height H above the top of the ramp at the highest point.

Problem 7 The figure below shows a pendulum of length L = 3.5 m. It is held so that the pendulum makes an angle θo = 35o and then released from rest. Find the speed of the pendulum when it is at its lowest point.

Problem 8 A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of r = 36 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the crest of the second hill.

Problem 9 A roller coaster (375 kg) moves from A (5.00 m above the ground) to B (20.0 m above the ground). Two nonconservative forces are present: friction does –2.00 × 104 J of work on the car, and a chain mechanism does +3.00 × 104 J of work to help the car up a long climb. What is the change in the car’s kinetic energy, ∆KE = KEf – KEo, from A to B?

Problem 10 A pitcher throws a 0.140 kg baseball, and it approached the bat at a speed of 40.0 m/s. The bat does Wnc = 70.0 J of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.

Problem 11 A car accelerates uniformly from rest to 20.0 m/s in 5.6 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 9.0 × 103 N, and (b) the weight of the car is 1.4 × 104 N.

Problem 12 The cheetah is one of the fastest accelerating animals, because it can go from rest to 27 m/s (about 60 mi/h) in 4.0 s. If its mass is 110 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower.