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“He had the odd feeling of being like a man in the act of adultery who is surprised when the woman's husband wanders into the room, changes his trousers, passes a few idle remarks about the weather and leaves again.” The Hitchhiker's Guide to the Galaxy Next Test will cover: Chapter 10: Springs and elasticity Chapter 11: Fluids and pressure Chapter 16: Waves and sound How do you think you scored on Test 2? A) 13 or more B) 10-12 C) 7-9 D) 4-6 E) <4 (really?!) Chapter 10: Springs. Please READ THE ENTIRE CHAPTER. Springs have an equilibrium length. To compress or stretch a spring requires force. Chapter 10: Springs. springs have an equilibrium length. To compress or stretch a spring requires force. Once stretched or compressed, a spring tries to return to the equilibrium length/position. Chapter 10: Springs. springs have an equilibrium length. To compress or stretch a spring requires force. The amount of force required to stretch/compress a spring is directly proportional to how far from equilibrium it has moved, and the stiffness of the spring. This is called Hooke's Law. Chapter 10: Springs. springs have an equilibrium length. To compress or stretch a spring requires force. Fx = -kx where k is called the spring constant and is a measure of a spring's stiffness. Chapter 10: Springs. springs have an equilibrium length. To compress or stretch a spring requires force. Fx = -kx where k is called the spring constant and is a measure of a spring's stiffness. Notice that the force always acts opposite to how the spring is moved. Chapter 10: Springs. Fx = -kx where k is called the spring constant. Example: A spring with k=3.67 N/m is stretched by 0.4 m. How much force does it take to hold the spring in this position? A) 0.4N B) -0.4N C) 1.5N D) -1.5N Chapter 10: Springs. Fx = -kx where k is called the spring constant. Example: A spring with k=3.67 N/m is stretched by 0.4 m. How much force does it take to hold the spring in this position? Fx = -(-kx) = 3.67(0.4) = 1.47 N which is C)1.5N Why wasn't it -1.5N? Chapter 10: Springs. Fx = -kx where k is called the spring constant. Example: A spring with k=3.67 N/m is stretched by 0.4 m. How much force does it take to hold the spring in this position? Fx = -(-kx) = 3.67(0.4) = 1.47 N which is C)1.5N Why wasn't it -1.5N? Spring Me holding the spring in place. Chapter 10: Springs. Fx = -kx where k is called the spring constant. Example: a spring with a 0.1N weight is at equilibrium. When an additional 0.1N weight is added, the spring stretches by 4cm. What is the spring constant? How would you do this problem? Chapter 10: Springs. Fx = -kx where k is called the spring constant. Example: a spring with a 0.1N weight is at equilibrium. When an additional 0.1N weight is added, the spring stretches by 4cm. What is the spring constant? k = |F/x| (spring constants are always positive) k = 0.1 / 0.04 = 2.5 N/m Springs: Potential Energy When a mousetrap is set, the energy within the spring is just waiting to be released, converting its energy to kinetic energy of the wire in the trap. That amount of energy is PE = ½kx2 where again x is how far displaced from equilibrium the spring is. Springs: Potential Energy PE = ½kx2 What is the potential energy stored in a typical mousetrap if the radius of the moving part is 0.0814 m (3.2 inches), and the spring constant (k) is 96.5 N/m? Can you do this problem? (Hint: Think about what X is) A) 0.320J C) 3.16J B) 0.639J D) 12.35J Springs: Potential Energy PE = ½kx2 What is the potential energy stored in a typical mousetrap if the radius of the moving part is 0.0814 m (3.2 inches), and the spring constant (k) is 96.5 N/m? Can you do this problem? PE = ½ kx2 in this case 'x' is the arc length the trap moves in its motion. It covers p radians, so X = s = rq = 0.0814(p) = 0.256 m. Springs: Potential Energy PE = ½kx2 What is the potential energy stored in a typical mousetrap if the radius of the moving part is 0.0814 m (3.2 inches), and the spring constant (k) is 96.5 N/m? Can you do this problem? PE = ½ kx2 with x = s = rq = 0.256 m. PE = ½(96.5)(0.2562) = 3.16 J This is C) 3.16J Springs: Potential Energy PE = ½kx2 Example problem: A child's toy is made to shoot a ping-pong ball (m=2.7 g) out of a tube. When the ball is loaded, the spring (k=18N/m) is stretched 9.5cm inside the tube. If the toy is shot straight up (neglecting any friction/air resistance), how high will the ball go? Can you do this problem? A) Yes B) No Springs: Potential Energy PE = ½kx2 m=2.7x10-3 kg x= 0.095m k=18N/m friction = 0, ho= 0, hf=? This is an energy conservation/conversion problem. Converting the spring's potential energy into gravitational potential energy: PES= PEg A) 1.2m B) 2.1m C) 3.0m D) 4.5m Springs: Potential Energy PE = ½kx2, m=2.7x10-3, kg x= 0.095m, k=18N/m, friction = 0, ho= 0, hf=? Energy conservation. Converting the spring's potential energy into gravitational potential energy: PES= PEg ½ kx2= mgh Did you get this far? Springs: Potential Energy PE = ½kx2, m=2.7x10-3, kg x= 0.095m, k=18N/m, friction = 0, ho= 0, hf=? Can you do this problem? Converting the spring's potential energy into gravitational potential energy: PES= PEg ½ kx2= mgh ½ (18)(0.0952) = (2.7x10-3)(9.8)hf 0.0812 = 0.0265hf → hf= 3.06 m C) 3.0m A problem for you tonight: A block (m=5.0kg) is released from point A and it slides down the incline where the coefficient of kinetic friction is 0.3. It hits a spring with a spring constant of 500 N/m. While it is being acted upon by the spring, assume it is on a frictionless surface. a) How far does the block go up the plane on the rebound from the spring? b) How far is the spring compressed? Answer revealed on Monday after spring break. Since W(ork) = DE, if we can change the energy of a spring system, we have done work. W = ½ k(xf - xo)2 W = ½ k(xf – xo)2 How much work does it take to stretch a spring 20cm if it has a spring constant of 545N/m? W = ½ k(xf – xo)2 How much work does it take to stretch a spring 20cm if it has a spring constant of 545N/m? W= ½ (545)(0.2-0.0)2 W= 10.9J If you take a block attached to a spring on a frictionless surface and you compress it, what's going to happen? If you take a block attached to a spring on a frictionless surface and you compress it, what's going to happen? The block moves back to the equilibrium point, but will it stop when it gets there? If you take a block attached to a spring on a frictionless surface and you compress it, what's going to happen? The block moves back to the equilibrium point, but will it stop when it gets there? No, it will move past that point, after which the spring force will act in the opposite direction. If you take a block attached to a spring on a frictionless surface and you compress it, what's going to happen? The block moves back to the equilibrium point, but will it stop when it gets there? No, it will move past that point, after which the spring force will act in the opposite direction. If the surface is truly frictionless, it will continue moving back-andforth forever! Simple Harmonic Motion Simple Harmonic Motion: the mass bounces back and forth. If you plot its position with time, you get the graph shown. Simple Harmonic Motion: the mass bounces back and forth. If you plot its position with time, you get the graph shown. A is the amplitude: this is the maximum distance (displacement) away from equilibrium the mass will move. Simple Harmonic Motion: the mass bounces back and forth. If you plot its position with time, you get the graph shown. A = amplitude. T is the time (in seconds) it takes to complete one period (down, up and back to the equilibrium position). This is the same as 1/frequency (units are Hertz, which is 1/seconds). 10 sec = 0.1Hz Simple Harmonic Motion: the mass bounces back and forth. If you plot its position with time, you get the graph shown. A = amplitude. T = time f = frequency. w is the angular frequency (not the angular velocity we had before (the relation to what we had before is wf= 2p.wv ) Simple Harmonic Motion: the mass bounces back and forth. If you plot its position with time, you get the graph shown. A = amplitude. T = time f = frequency. w = angular frequency. The position at any time can also be determined. And there are relations to the spring constant and mass. Example: A ball (m=0.2kg) is bouncing (A=6cm) on a spring (k=15N/m) and crosses equilibrium at t=0. Where is it 12.3 seconds later? Can you solve this problem? Example: A ball (m=0.2kg) is bouncing (A=6cm) on a spring (k=15N/m) and crosses equilibrium at t=0. Where is it 12.3 seconds later? Can you solve this problem? Use the second part of the y equation. Example: A ball (m=0.2kg) is bouncing (A=6cm) on a spring (k=15N/m) and crosses equilibrium at t=0. Where is it 12.3 seconds later? Can you solve this problem? Use the second part of the y equation. y = A sin[(k/m)1/2]t = 0.06 sin[(15/0.2)1/212.3] Example: A ball (m=0.2kg) is bouncing (A=6cm) on a spring (k=15N/m) and crosses equilibrium at t=0. Where is it 12.3 seconds later? Can you solve this problem? Use the second part of the y equation. y = A sin[(k/m)1/2t] = 0.06 sin[(15/0.2)1/212.3] = 0.06*sin(8.66*12.3) = 0.06(-0.29) = -0.0175 m