AP Physics Chapter 5 Work and Energy

Chapter 5: Work and Energy 5.1 Work Done by a Constant Force 5.2 Work Done by a Variable Force 5.3 The Work-Energy Theorem: Kinetic Energy 5.4 Potential Energy 5.5 The Conservation of Energy 5.6 Power

Chapter 5: Learning Objectives Work and the Work-energy Theorem Students will understand the definition of work, including when it is positive, negative, or zero, so they can: Calculate the work done by a specified constant force on an object that undergoes a specified displacement. Relate the work done by a force to the area under a graph of force as a function of position, and calculate this work in the case where the force is a linear function of position. Use the scalar product operation to calculate the work performed by a specified constant force F on an object that undergoes a displacement in a plane.

Chapter 5: Learning Objectives Work and the Work-energy Theorem Students will understand and be able to apply the work-energy theorem, so they can: Calculate the change in kinetic energy or speed that results from performing a specified amount of work on an object. Calculate the work performed by the net force, or by each of the forces that make up the net force, on an object that undergoes a specified change in speed or kinetic energy. Apply the theorem to determine the change in an object’s kinetic energy and speed that results from the application of specified forces, or to determine the force that is required in order to bring an object to rest in a specified distance.

Chapter 5: Learning Objectives Forces and Potential Energy Students will understand the concept of potential energy, so they can: Write an expression for the force exerted by an ideal spring and for the potential energy of a stretched or compressed spring. Calculate the potential energy of one or more objects in a uniform gravitational field.

Chapter 5: Learning Objectives Conservation of Energy Students will understand the concept of mechanical energy and of total energy, so they can: Describe and identify situations in which mechanical energy is converted to other forms of energy. Analyze situations in which an object’s mechanical energy is changed by friction or by a specified externally applied force. Students will understand conservation of energy, so they can: Identify situations in which mechanical energy is or is not conserved. Apply conservation of energy in analyzing motion of systems of connected objects, such as an Atwood’s machine. Apply conservation of energy in analyzing the motion of objects that move under the influence of springs.

Chapter 5: Learning Objectives Power Students will understand the definition of power, so they can: Calculate the power required to maintain the motion of an object with constant acceleration (e.g., to move an object along a level surface, to raise an object at a constant rate, or to overcome friction for an object that is moving at a constant speed.) Calculate the work performed by a force that supplies constant power, or the average power supplied by a force that performs a specified amount of work.

Homework for Chapter 5 Read Chapter 5 HW 5.A : p. 165 3, 6-11, 16. HW 5.B: p. 167: 19-26, 28, 31. HW 5.C: pp.167-168: 32, 34, 38-40, 45-52. HW 5.D: p.168: 60-64, 66, 67, 70. HW 5.E: p.170: 72,73,75-81.

5.1 Work Done by a Constant Force

Warmup: Work, Work, Work! Physics Daily WarmUps # 76 The quantity work is defined as the product of the force applied to an object multiplied by the distance through with the force is applied. This means that if no displacement of the object occurs, no work is done on the object even though the force applied may be quite large. Reference to the work we do is a large part of our daily conversation. We all have opinions as to which jobs require more of less work. However, this common usage of the word work does not always match up with the physics definition. Use the column on the left to arrange the list of jobs in order of hardest work to easiest in your opinion. Use the column on the right to arrange them in order of most work to least work according to physics. Opinion (hardest to easiest) Jobs Physics (most to least) (hardest) store clerk accountant package delivery driver (easiest) furniture mover furniture mover package delivery driver store clerk accountant

5.1 Work Done by a Constant Force Definition: Work done by a constant force is equal to the product of the magnitudes of the displacement and the component of the force parallel to the displacement. W = (F cosӨ)d, where F is the magnitude of the force vector d is the magnitude of the displacement vector Ө is the angle between the two vectors (Warning – this angle is not necessarily measured from the horizontal) When the angle is zero, cos Ө = 1, and W = F·d When the angle is 180°, cos Ө = -1 and W = - F·d (yes, negative work!) example: the force of brakes to slow down a car. Although force and displacement are vectors, work is a scalar quantity. The SI unit of work is the N·m, which is called a joule (J).

5.1 Work Done by a Constant Force

5.1 Work Done by a Constant Force If there is no displacement, no work is done: W = 0. For a constant force in the same direction as the displacement, W = Fd. For a constant force at an angle to the displacement, W = (FcosӨ) d

5.1 Work Done by a Constant Force Example 1: A 500 kg elevator is pulled upward by a constant force of 5,500 N for a distance of 50.0 m. a) Find the work done by the upward force. b) Find the work done by the gravitational force. c) Find the work done by the net force (the net work done on the elevator). Solution: Given: Fup = 5,500 N w = mg = (500 kg)(9.80 m/s2) = 4,900 N d = 50.0 m Unknown: a) Wup b) Wgrav c) Wnet a) The displacement is upward and the upward force is (of course) upward, so the angle between them is zero. Therefore, Wup = (F cos Ө) d = (Fup cos 0°) d = (5,500 N)(1)(50.0m) = 2.75 x 105 J The angle between displacement and the gravitational force (weight vector) is 180°. So, Wgrav = (w cos 180°) d = (4,900 N)(-1)(50.0 m) = -2.45 x 105 J The work done by the net force is equal to the net work done on the elevator. Wnet = Wup + Wgrav = 2.75 x 105 J + (-2.45 x 105 J) = 3.0 x 104 J Note: Wnet is also equal to Wnet = (Fnet cos Ө) d, where Fnet = F - w Fup w (weight) d

5.1 Work Done by a Constant Force The work done by a force is equal to the area under the curve in a force vs. position graph. Example 2: A force moves an object in the direction of the force. The graph shows the force versus the object’s position. Find the net work done when the object moves a) from 0 to 2.0 m b) from 2.0 to 4.0 m c) from 4.0 to 6.0 m d) from 0 to 6.0 m Solution: Work done is equal to the area under the curve. The area under the curve from 0 to 2.0 m is a triangle. The area is ½ (base x height), so W0-2 = ½ (2.0 m - 0)(20 N) = 20 J The area under the curve is W2-4 = (4.0 m – 2.0 m)(20 N) = 40 J The area under the curve is W4-6 = ½ (6.0 m – 4.0 m)(20 N) = 20 J The sum of the areas is W0-2 + W2-4 + W4-6 = 20 J + 40 J + 20 J = 80 J F (N) 30 20 10 0.0 1.0 2.0 3.0 4.0 5.0 6.0 s (m)

5.1 Work Done by a Constant Force: Check for Understanding

5.1 Work Done by a Constant Force: Check for Understanding

5.1 Work Done by a Constant Force: Check for Understanding

5.1 Work Done by a Constant Force: Check for Understanding

5.1 Work Done by a Constant Force: Check for Understanding

5.1 Work Done by a Constant Force: Check for Understanding

Homework for Section 5.1 HW 5.A : p. 165 3, 6-11,16.

5.2 Work Done by a Variable Force

5.2 Work Done by a Variable Force Forces that do work are not always constant. example: To move a sofa, you push harder and harder until you can overcome static friction. example: Stretching or compressing a spring. As the spring gets stretched (or compressed) farther and farther, the restoring force of the spring (the force that opposes stretching) gets greater and greater. The applied force is directly proportional to the change in length of the spring from its unstretched length. The spring constant, k, describes the stiffness of the spring. The resulting equation is known as Hooke’s Law: Fs = -kx The minus sign indicates the spring force is opposite to the direction of displacement.

Spring Force 5.2 Work Done by a Variable Force An applied force F stretches the spring, and the spring exerts an equal and opposite force Fs on the hand. The magnitude of the force depends on the change in the spring’s length. This is often referenced to the position of a mass on the end of the spring.

5.2 Work Done by a Variable Force The work done by an external force in stretching or compressing a spring (overcoming the spring force) is W = ½ kx2 where x is the stretch or compression distance and k is the spring constant. F F = kx work = area under the curve area of the triangle = ½ (base x height) work = ½ (x)(kx) = ½ kx2 slope = k x

5.2 Work Done by a Variable Force The reference position xo is chosen for convenience. xo may be chosen to be at the end of the spring in its unloaded position. b) xo may be at the equilibrium position when a mass is suspended on a spring. This is convenient when the mass oscillates up and down on the spring.

5.2 Work Done by a Variable Force Example: A spring of spring constant 20 N/m is to be compressed by 0.10 m. What is the maximum force required? What is the work required? Solution: Given: k = 20 N/m x = -0.10 m Unknown: a) Fs(max) b) W From Hooke’s law, the maximum force corresponds to the maximum compression. Fs(max) = -kx = -(20 N/m)(-0.10 m) = 2.0 N b) W = ½ kx2 = ½ (20 N/m)(-0.10 m)2 = 0.10 J

5.2 Work Done by a Variable Force: Check for Understanding 4 3 2 1

5.2 Work Done by a Variable Force

Homework for Section 5.2 HW 5.B: p. 167: 19-26, 28, 31.

The Work Energy Theorem 5.3 Kinetic Energy 5.4 Potential Energy

5.3 The Work-Energy Theorem: Kinetic Energy v = velocity

5.3 The Work-Energy Theorem: Kinetic Energy

5.3 The Work-Energy Theorem: Kinetic Energy Example: An object hits a wall and bounces back with half of its original speed. What is the ratio of the final kinetic energy to the initial kinetic energy? Solution: Given: v = vo/2 Unknown: K/Ko Ko = 1/2 mvo2 and K = ½ mv2 = ½ m (vo/2)2 = ¼ (1/2 mvo2) So, K/ Ko = 1/4

5.3 The Work-Energy Theorem: Kinetic Energy The work-energy theorem offers a convenient way to solve kinematics and dynamics problems. Example: The 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? Given: m = 60.0 kg vo = 25.0 m/s v = 0 fk = 50.0 N Unknown: d The kinetic friction force fk is the only unbalanced force and the angle between the friction force and the displacement is 180°. Wnet = (FcosӨ) d = (fk cos 180°) d = (50.0 N)(-1) d and, Wnet = K- Ko, = ½ mv2 – ½ mvo2 = ½(60.0 kg)(0)2 – ½ (60.0 kg)(25.0 m/s)2 Solving, d = 3.75 x 102 m

5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding

5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding

5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding

5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding

5.4 The Work-Energy Theorem: Potential Energy

5.4 The Work-Energy Theorem: Potential Energy

5.4 The Work-Energy Theorem: Potential Energy The work done in lifting an object is equal to the change in gravitational potential energy. W = F∆ h = m g (h – ho)

5.4 The Work-Energy Theorem: Potential Energy Kinetic and Potential Energy Throwing a ball in the air converts its kinetic energy to potential energy, and back again.

5.4 The Work-Energy Theorem: Potential Energy Reference Point and Change in Potential Energy The reference point (zero height) you choose may give rise to negative potential energy. This is call a potential energy well. The well may be avoided by choosing a new reference point.

5.4 The Work-Energy Theorem: Potential Energy

5.4 The Work-Energy Theorem: Potential Energy

5.4 The Work-Energy Theorem: Potential Energy Example 5.6: A 10.0 kg object is moved from the second floor of a house 3.00 m above the ground to the first floor 0.30 m above the ground. What is the change in gravitational potential energy?

5.4 The Work-Energy Theorem: Potential Energy Example 5.7: A spring with a spring constant of 15 N/m is initially compressed by 3.0 cm. How much work is required to compress the spring an additional 4.0 cm?

5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding

5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding

5.4 The Work-Energy Theorem: Potential Energy

5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding

5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding

5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding Or, use opp = hyp (sin 30) = 0.5 m

Homework for Sections 5.3 & 5.4 HW 5.C: pp.167-168: 32, 34, 38-40, 45-52.

5.5 The Conservation of Energy

Warmup: Road Hazard

5.5 The Conservation of Energy

5.5 The Conservation of Energy Remember, mechanical energy is kinetic or potential. If there is a nonconservative force such as friction doing work on the system, the total mechanical energy of the system is not conserved, but total overall energy is conserved.

5.5 The Conservation of Energy Example 5.8: A 70 kg skier starts from rest on the top of a 25 m high slope. What is the speed of the skier on reaching the bottom of the slope? (Neglect friction)

5.5 The Conservation of Energy If there is a nonconservative force doing work in a system, the total mechanical energy of the system is not conserved. However, the total energy (not mechanical!) of the system is still conserved. Some of the total energy is used to overcome the work done by the nonconservative force. The difference in mechanical energy is equal to the work done by the nonconservative force, that is Wnc = E – Eo = E

5.5 The Conservation of Energy Example 5.10: In Example 5.8, if the work done by the kinetic friction force is -6.0 x 103 J (the work done by kinetic friction force is negative because the angle between the friction force and the displacement is 180˚). What is the speed of the skier at the bottom of the slope?

5.5 The Conservation of Energy Example 5.9: A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal spring with spring constant 2.0 x 106 N/m. What is the maximum compression of the spring? (Neglect the mass of the spring.)

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

5.5 The Conservation of Energy: Check for Understanding

Explanation of the Math for Pulley Problem: 5.5 The Conservation of Energy Explanation of the Math for Pulley Problem: Uo = K Just block A: K = ½ m vA2 Both blocks together: K = ½ 3m vAB2 Since the change in energy is the same for both cases, set them equal to each other: ½ 3m vAB2 = ½ m vA2 vAB2 = vA2/ 3 vAB = vA / √3

Homework for Section 5.5 HW 5.D: p.168: 60-64, 66, 67, 70.

5.6 Power

Warmup: Rock-It

5.6 Work

5.6 Power A common British unit of power is the horsepower (hp) and 1 hp = 746 W.

5.6 Power

5.6 Power

5.6 Power Example 5.11: A 1500 kg car accelerates from 0 to 25 m/s in 7.0 s. What is the average power delivered to the car by the engine? Ignore all frictional and other losses.

5.6 Power: Check for Understanding

5.6 Power: Check for Understanding

5.6 Power: Check for Understanding

5.6 Power: Check for Understanding

Homework for Section 5.6 HW 5.E: p.170: 72-73, 75-81.

Chapter 5 Formulas K = ½ mv2 kinetic energy Ug = mgh gravitational potential energy W = F r cos  work Pave = W power as defined by work over time t P = Fv cos  power if the force is making an angle  with v Fs = -kx Hooke’s Law; relates spring force with spring constant and change in length Us = ½ kx2 elastic potential energy