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Physics in Context Unit 7 Work and Energy 3 Key Objectives State the equation for calculating the elastic potential energy of a spring, and solve problems using this equation. Solve problems involving the inclined plane as a simple machine. List other types of simple machines. 7.3 ENERGY Elastic Potential Energy and Springs • Consider the arrangement shown in the diagram: 7.3 ENERGY Elastic Potential Energy and Springs • Here a spring is attached to a wall and to a mass resting on a horizontal, fric tionless table. 7.3 ENERGY Elastic Potential Energy and Springs • If we apply a force on the mass and displace it to the right, we have done work. 7.3 ENERGY Elastic Potential Energy and Springs • This work has been converted into the spring’s potential energy, a quantity we call elastic potential energy. 7.3 ENERGY Elastic Potential Energy and Springs • We can calculate the work done in stretching the spring as follows. 7.3 ENERGY Elastic Potential Energy and Springs • We know that springs obey Hooke’s law, a graph of which is shown below. 7.3 ENERGY Elastic Potential Energy and Springs • Since the area under this graph equals the work done, and we know that the area of a triangle is equal to 1/2 (base height), we use the equations 7.3 ENERGY Elastic Potential Energy and Springs • Then we have 7.3 ENERGY Elastic Potential Energy and Springs • and the potential energy of the spring (PEs) is given by: 7.3 ENERGY Elastic Potential Energy and Springs • PROBLEM • A spring whose constant is 2.0 newtons per meter (k) is stretched 0.40 meter (x) from its equilibrium position. • What is the increase in the elastic potential energy (∆PEs) of the spring? 7.3 ENERGY Elastic Potential Energy and Springs • • • • SOLUTION ∆PEs = ½ kx2 ∆PEs = ½ (2.0 N/m)(0.40m)2 ∆PEs = 0.16 J Assessment Question 1 • A spring whose constant is 1.2 N/m (k) is stretched 0.65 m (x) from its equilibrium position. • What is the increase in the elastic potential energy (∆PEs) of the spring? • ∆PEs = ½ kx2 = ½ (1.2 N/m)(0.65 m)2 = A.0.25 J B.1.6 J C.4.7 J D.20 J 7.3 ENERGY Elastic Potential Energy and Springs • SOLUTION • What would happen if we released the spring after stretching it? • The force exerted on the mass by the spring (the restoring force) would displace the mass toward the wall (to the left). 7.3 ENERGY Elastic Potential Energy and Springs • SOLUTION • The mass would then overshoot its equilibrium position and compress the spring. • In turn, the spring would exert a, force on the mass away from the wall (to the right). 7.3 ENERGY Elastic Potential Energy and Springs • SOLUTION • Since friction is absent, this back-and-forth motion, namely, SHM would continue indefinitely. 7.3 ENERGY Elastic Potential Energy and Springs • SOLUTION • As the spring moved back and forth, there would be a continual exchange between kinetic and potential energies, as shown in the diagram. 7.3 ENERGY Elastic Potential Energy and Springs • SOLUTION Assessment Question 2 If we released the spring after stretching it. All of the following are true EXCEPT? A. The force exerted on the mass by the spring (the restoring force) would displace the mass toward the wall. B. The mass would then overshoot its equilibrium position and compress the spring, then the spring would exert a, force on the mass away from the wall (to the right). C. If friction is absent, this back-and-forth motion, namely, Simple Harmonic Motion would continue indefinitely. D. As the spring moved back and forth, the kinetic and potential energies remain constant. 7.3 ENERGY Elastic and Inelastic Collisions • Imagine a ball bouncing repeatedly on a sidewalk, as illustrated below. 7.3 ENERGY Elastic and Inelastic Collisions • After each successive bounce, the ball’s height above the ground diminishes; eventually the ball comes to rest on the ground. 7.3 ENERGY Elastic and Inelastic Collisions • When the ball is on the ground, part of its kinetic energy is lost and there is an incomplete conversion to potential energy, a phenomenon known as a partially inelastic collision. 7.3 ENERGY Elastic and Inelastic Collisions • If the ball stuck to the ground after its first bounce, the collision would be termed totally inelastic. 7.3 ENERGY Elastic and Inelastic Collisions • If, however, the ball rose repeatedly to the same height, the collisions would be termed elastic. Assessment Question 3 If a ball is bouncing repeatedly on a sidewalk. All of the following are true EXCEPT? A. After each successive bounce, part of its kinetic energy is lost and there is an incomplete conversion to potential energy and the ball’s height above the ground diminishes and the ball comes to rest on the ground which is known as a partially inelastic collision. B. If the ball stuck to the ground after its first bounce, the collision would be termed totally inelastic collision. C. If the ball rose repeatedly to the same height, the collisions would be termed elastic collisions. D. If the ball increased its height after each bounce this would be termed a superelastic collision. 7.3 ENERGY Elastic and Inelastic Collisions • In an elastic collision both kinetic energy and momentum are conserved, as outlined below: Assessment Question 4 • Two objects of the same mass moving at 10 m/s (vi) in opposite directions collide with 500 J (KEi) of force and stop on impact. What is the mass of each object? • KE1i + KE2i = KE1f + KE2f • ½ mv1i2 + ½ mv2i2 = 500 J • ½ m∙(10 m/s)2 + ½ m∙(10 m/s)2 = 500 J • (100 m2/s2) m = 500 J m = 500 J / 100 m2/s2 A.5 kg C. 2500 kg B.100 kg D. 50000 kg 7.3 ENERGY Elastic and Inelastic Collisions • In an elastic collision both kinetic energy and momentum are conserved, as outlined below: 7.3 ENERGY Elastic and Inelastic Collisions • In inelastic collisions, the kinetic energy that is “lost” is converted into internal energy (Q) (heat) of the objects by frictional forces. 7.3 ENERGY Elastic and Inelastic Collisions • In a non ideal mechanical system, the total energy is constant. 7.3 ENERGY Elastic and Inelastic Collisions • This form of energy will be discussed in Chapter 8. • Systems upon which frictional or other external forces act are called non ideal mechanical systems Assessment Question 5 If a ball is bouncing repeatedly on a sidewalk. All of the following are true EXCEPT? A. In an elastic collision both kinetic energy and momentum are conserved. B. In inelastic collisions, the kinetic energy that is “lost” is converted into internal energy (heat) (Q) of the objects by frictional forces. C. In a non ideal mechanical system, the final total energy is greater than the initial total energy. D. Systems upon which frictional or other external forces act are called non ideal mechanical systems 7.3 ENERGY Elastic and Inelastic Collisions • The equation representing the laws of conservation of energy on page 129 can be expanded to include Q, and the total energy (ET) is given by Assessment Question 6 • An object with a mass of 5 kg (m) falls down the side of a cliff from a distance of 10 m (h) from the ground at 10 m/s (v). g = 9.8 m/s2 • If the total energy is 1000 J (ET), what is the energy lost to heat (Q)? • ET = PE + KE + Q • Q = ET - (PE + KE) = ET – (mgh + ½ mv2) Q = 1000 J -[(5 kg∙9.8 m/s2∙10 m) +½(5 kg∙(10 m/s))2] A. 55 J C. 260 J B. 140 J D. 530 J 7.4 SIMPLE MACHINES AND WORK • A simple machine is a device that allows work to be done and offers an advantage to the user. • We need to use machines because the force available to us is not always adequate. 7.4 SIMPLE MACHINES AND WORK • For example, we cannot lift a 1000-newton box using our muscles alone. • However, a simple machine such as a pulley or an inclined plane may allow us to accomplish this task. 7.4 SIMPLE MACHINES AND WORK • Suppose we used a system of pulleys and exerted a downward force of 250 newtons to raise the 1000-newton box. • We say that the pulley offers us a mechanical advantage of 4. 7.4 SIMPLE MACHINES AND WORK • We calculate the mechanical advantage from this relationship: (Fout /Fin) • In nature, however, one never gets something for nothing and we pay for this mechanical advantage by pulling the rope of the pulley four times as far as the box is lifted. 7.4 SIMPLE MACHINES AND WORK • Since energy must be conserved, we could have deduced this fact from the ideal relationship that work input must equal work output: 7.4 SIMPLE MACHINES AND WORK • The relationship given above assumes that no friction is present in the machine, and the mechanical advantage is called the ideal mechanical advantage (IMA). 7.4 SIMPLE MACHINES AND WORK • In reality, all machines have friction, so the actual mechanical advantage (AMA) is always less than the IMA. • The IMA is based on the relative distances involved in the operation of the machine (din /dout) while the AMA is based on the ratio Fout / Fin . Assessment Question 7 All of the following are true EXCEPT? A. A simple machine is a device that allows work to be done and offers an advantage to the user. B. If a person uses a system of pulleys and exerted a downward force of 250 newtons to raise the 1000-newton box, the pulley system offers a mechanical advantage of 4 and the person pulls the rope ¼ the distance that the box is lifted. C. If no friction is present in the machine, the mechanical advantage is called the Ideal Mechanical Advantage (IMA). In reality, all machines have friction, so the Actual Mechanical Advantage (AMA) is always less than the Ideal Mechanical Advantage IMA. D. The Ideal Mechanical Advantage (IMA) is based on the relative distances involved in the operation of the machine (din /dout) while the Actual Mechanical Advantage (AMA) is based on the ratio Fout / Fin . We calculate the mechanical advantage from the relationship: (Fout /Fin) 7.4 SIMPLE MACHINES AND WORK • PROBLEM • A 100-newton object is moved 2 meters up an inclined plane whose end is lifted 0.5 meter from the floor. • If a force of 50 newtons is needed to accomplish this task, calculate the (a) IMA, (b) AMA, and (c) efficiency of the inclined plane. 7.4 SIMPLE MACHINES AND WORK • SOLUTION • A diagram will help us to identify the relevant quantities: 7.4 SIMPLE MACHINES AND WORK • SOLUTION • The input force (Fin) is the force needed to move the object along the plane (50 N). 7.4 SIMPLE MACHINES AND WORK • SOLUTION • The output force (Fout) is the force that has been lifted (100 N). 7.4 SIMPLE MACHINES AND WORK • SOLUTION • The input distance (din) is the distance the object is moved along the plane (2 m). 7.4 SIMPLE MACHINES AND WORK • SOLUTION • The output distance (dout) is the distance to which the object has been raised (0.5 m). 7.4 SIMPLE MACHINES AND WORK • SOLUTION • The output distance (dout) is the distance to which the object has been raised (0.5 m). 7.4 SIMPLE MACHINES AND WORK • PROBLEM • A 100-newton object is moved 2 meters up an inclined plane whose end is lifted 0.5 meter from the floor. • If a force of 50 newtons is needed to accomplish this task, calculate the (a) IMA, 7.4 SIMPLE MACHINES AND WORK • SOLUTION • a) calculate the Ideal Mechanical Advantage (IMA) Assessment Question 8 • A 150 N (Fout) object is moved 15 m (din) up an inclined plane whose end is lifted 2.5 m (dout) from the floor. • If a force of 50 N (Fin) is needed to accomplish this task, calculate the Ideal Mechanical Advantage (IMA) • IMA = din/dout = 15 m / 2.5 m = A.0.5 C. 3 B.1 D. 6 7.4 SIMPLE MACHINES AND WORK • PROBLEM • A 100-newton object is moved 2 meters up an inclined plane whose end is lifted 0.5 meter from the floor. • If a force of 50 newtons is needed to accomplish this task, calculate the (b) AMA 7.4 SIMPLE MACHINES AND WORK • SOLUTION • b) calculate the Actual Mechanical Advantage (AMA) Assessment Question 9 • A 150 N (Fout) object is moved 15 m (din) up an inclined plane whose end is lifted 2.5 m (dout) from the floor. • If a force of 50 N (Fin) is needed to accomplish this task, calculate the Actual Mechanical Advantage (AMA) AMA = Fout/Fin = 150 N / 50 N = A. 0.5 C. 3 B. 1 D. 6 7.4 SIMPLE MACHINES AND WORK • PROBLEM • A 100-newton object is moved 2 meters up an inclined plane whose end is lifted 0.5 meter from the floor. • If a force of 50 newtons is needed to accomplish this task, calculate the (c) efficiency of the inclined plane. 7.4 SIMPLE MACHINES AND WORK • SOLUTION • calculate the (c) efficiency of the inclined plane. 7.4 SIMPLE MACHINES AND WORK • The efficiency of the machine is the AMA/IMA ratio; this value is always less than 100%. • Examples of simple machines include inclined planes, pulleys, levers, wheels and axles, and screwdrivers. Conclusion • An elastic collision is one in which momentum and kinetic energy are conserved. • When gas molecules collide with the walls of a container, these collisions are very nearly elastic. • Simple machines provide examples of how work can be used to a person’s advantage in performing such chores as lifting heavy objects and exerting large forces that can accomplish a task such as cutting through steel. Assessment Question 10 • A 150 N (Fout) object is moved 15 m (din) up an inclined plane whose end is lifted 2.5 m (dout) from the floor. • If a force of 50 N (Fin) is needed to accomplish this task, calculate Efficiency: Efficiency = AMA / IMA = (Fout/Fin) / (din/dout) Efficiency = (150 N / 50 N) / (15 m / 2.5 m) = A. 0.25 B. 0.5 C. 0.75 D. 1