PSC 151 - Lab 6

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Physical Science Laboratory: PSC 151 Lab 6 Rev 12/19/18 Mod: 2/9/19 VPL Lab – Dynamics, Energy and Work - mod 1 Name: Lab 6 Conservation of Mechanical Energy; Work Date: P URPOSE  To study conservation of Mechanical Energy for a cart moving along an incline  To observe the scalar nature of energy  To examine the non-conservative nature of the friction force. T HEORY The kinetic energy of a body is the energy associated with its motion. Kinetic Energy, 𝑲𝑬 = ½ 𝒎𝒗 𝟐 (1) The potential energy of a physical system is the energy associated with the arrangement of the objects making up the system. This energy is due to the forces among the objects in the system and the work that those forces can do in changing the arrangement of the objects in the system. For example, moving a pair of magnets closer together or farther apart changes the amount of potential energy of the system. In the case of gravitational potential energy for an object near Earth, the energy is due to the force of gravity acting between the objects. Gravitational Potential Energy, 𝑷𝑬 = mg 𝒉 (2) If an object at some height above the earth is allowed to fall to a lower altitude, the loss in potential energy of the system of the earth and the object would be equal to the work done by the force of gravity. In such a case the motion of the earth is negligible and we usually say that it is the falling object that loses the energy. The change in potential energy of the object is given by ∆𝑷𝑬 = 𝑷𝑬 𝒇 − 𝑷= 𝒎𝒈𝒉 𝒇 − 𝒎𝒈𝒉 𝒐 (3) where h is measured relative to some arbitrary level where we set h = 0. We define the sum of the kinetic and potential energy of an object (or system) as its total mechanical energy . That is, ME = PE + KE (4) Note that energy is a scalar quantity, so v is the speed, which is always positive. And h is the height, which can have a negative sign if an object is below the h = 0 reference level. But this does not represent direction. A swinging pendulum is a good example of a system which undergoes a steady change in PE and KE . As the bob rises, its PE increases and its KE decreases. As it swings back downward the reverse occurs. If there are no forces other than gravity acting, the decrease in one type of energy is exactly matched by the increase in the other. Thus the total ME remains constant. If the work on the object by non-conservative forces such as friction is zero, the total mechanical energy will remain constant. For such cases we can say that mechanical energy is conserved . If Wnc = 0, PEo + KEo = PEf + KEf or Δ PE + Δ KE = 0 Principle of Conservation of Mechanical Energy A car moving up or down a hill is another example of energy being converted between KE and PE . For a real car the total mechanical energy would not remain constant however, since non-conservative forces such as friction are also involved.

Physical Science Laboratory: PSC 151 Lab 6 Rev 12/19/18 Mod: 2/9/19 VPL Lab – Dynamics, Energy and Work - mod 2 E QUIPMENT Dynamics Track PENCIL E XPLORE THE A PPARATUS Open the Dynamics Track. You should be familiar with most of the features of ramp and cart system from your study of kinematics and dynamics. There are just a few new features which you should familiarize yourself with now. Setting the initial velocity, vo Figures 1 and 3 show parts of the control panel beneath the right end of the track. These controls are used to set the initial velocity of the cart as well as the static and kinetic friction between the cart and track. The panel’s appearance changes according to the situation. Figure 1 – Initial Velocity Controls Clicking the Go→ button on the left end of the panel gives the cart an instantaneous velocity, vo . Vo can range from - 300 cm/s to +300 cm/s. You can change the direction of the velocity by inserting (or deleting) a – sign or use the Vo stepper gadget ( ) to move through the range of positive and negative velocities. Adjusting the static and kinetic friction, μks By default the cart runs on frictionless wheels. This is indicated by the selected radio button next to “wheels.” To allow you to investigate motion with frictional resistance the wheels can be replaced by a friction pad with variable friction coefficients, kinetic, µK and static, µS. When the unselected radio button is clicked it becomes selected and the friction pad controls become active as shown in Figure 3. Wheels Friction Pad Figure 2 – Cart Modes The two left steppers show the default values for µKinetic and µStatic. These steppers allow you to adjust these two values independently. Well, almost independently. µK must remain a respectful amount below µS. µKS? works similarly to Vo? That stepper allows you to choose from among several pairs of fixed but unknown (to you) µK and µS values. Figure 3 – Initial Velocity Controls P ROCEDURE I. Conservation of Mechanical Energy A. Using conservation of mechanical energy to predict the change in height, Δ h , of a cart traveling up a frictionless ramp and coming to a halt. Consider a cart traveling up a frictionless ramp. It will start with a certain initial velocity, vo , and a corresponding amount of initial kinetic energy, ½ mvo 2. Let’s examine this situation with our apparatus. Adjust the ramp angle to 5°. Set v o to 160 cm/s. Click the dynamic vectors icon, . Click Go→ and repeat as necessary to observe the following. Note the pink W x vector (shown as W(x)) which is constant for a given track angle and always points down the track. This is the component of the weight of the cart acting parallel to the track. As the cart moves, this force does positive or negative work on the cart depending on whether the cart is moving in the direction of the force or opposite it. The green velocity vector varies in magnitude and direction as the cart’s kinetic energy changes. So a change in the length of this vector would indicate a Δ KE .

o Physical Science Laboratory: PSC 151 Lab 6 VPL Lab – Dynamics, Energy and Work - mod 3 Rev 12/19/18 Mod: 2/9/19 When the cart goes up the ramp, Wx and Δ x are in opposite directions so negative work is done to the cart. Going down the ramp, positive work is done to the cart so,  Going up: Work = - | Wx Δx | so Δ KE is also negative, so KE decreases  Going down: Work = + | Wx Δx | so Δ KE is also positive, so KE increases  Up then down: Work = - | Wx Δx | + | Wx Δx | = 0, so Δ KE = 0 (starts and ends at 160 cm/s (speeds, not velocities)) A much more accessible way to describe this series of events involves replacing the work by gravity with the change in potential energy of the cart. Again the cart starts with some initial KE o and we’ll arbitrarily set its initial PE o to zero. As the cart loses KE going up the track, it gains an equal amount of PE . This is reversed on the way back down. PEo + KEo = PEf + KEf (5) mgho + ½ mv 2 = mghf + ½ mvf2 (6) To explore how these equations describe the motion of the cart you will A. Use conservation of energy to predict the change in height for a cart given an initial speed at the bottom of the ramp. B. Check your prediction by measuring its actual change in height geometrically. To make prediction of the change in height of the ramp we can use energy conservation: mg(h f – h o) = KE f – KE o.  mg Δh = KE f – KE o A. Predicted Δ h from Conservation of Mechanical Energy Table 1. Change in Height, Δ h , of the Cart on the Ramp Cart mass = 0.250 kg Ramp angle, θ = 5 ° Initial position of the Cart at the bottom of the ramp: x 0 = 0.074 m Final position of the Cart at the top of the ramp: x f = 1.92 m Initial velocity of Cart from bottom of ramp: vo = 1.78 m/s KE o = ½ mvo2 = (0.5)(0.250 kg)(1.78 m/s)2 = Final velocity of Cart at the top of the ramp: vf = 0 m/s KE f = ½ mvf2 = (0.5)(0.250 kg)(0 m/s)2 = Δh predicted (from energy conservation): mg Δh = KE f – KE o Δh = KE o /mg = -0.396 J/(0.250 kg)(9.81 m/s 2 ) =  mg Δh = 0 – KE o = KE 0  The initial kinetic energy, KE o , can be found from the empty cart’s mass and its initial speed which you’ll set using the initial velocity, Vo, control discussed above. 1. Set the ramp angle to any angle from +1° to +5°. Record your chosen ramp angle in Table 1. 2. Set recoil to 0 and leave it there for all parts of the lab. 3. Turn on the ruler. 4. By trial and error find an initial speed that will send the cart to approximately the 192 cm point. Record this as your initial velocity, vo in Table 1. Note that the cart will start at this initial velocity from a dead start when you click Go→. Yes, this would be impossible to do with real physical equipment.

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Physical Science Laboratory: PSC 151 Lab 6 VPL Lab – Dynamics, Energy and Work - mod 4 Rev 12/19/18 Mod: 2/9/19 Fig. 4a Cart starts from 0.074 m Fig. 4b Cart starts at 0.20 m Fig. 3 x-t Motion Graph showing that Cart reaches maximum distance of 1.92 m up the incline when v o = 1.78 m/s 5. Calculate and record KE o in joules for the .250-kg cart. (Careful with units!) 6. Turn on the brake, and drag the cart so that the cart mast is at approximately the 20 cm point on the ruler. 8. Turn on the sensor. 8. Quickly click Go→ to launch the cart. Turn off the sensor when cart returns to the bottom. Note: You’ll need data from the lab apparatus data table later so don’t use the sensor again until you complete Table 1. 9. Record the cart’s final velocity, and final kinetic energy, vf , and KE f , respectively, at the top of the cart’s travel. (There’s no data required for this. Just think about it.) 10. Using conservation of mechanical energy, calculate and record the cart’s predicted change in height, Δ h . ( hf – ho ) Show calculations of your predicted (theoretical) Δh here. According to mechanical energy conservation, the total energy at the top of the ramp = the total energy at the bottom of the ramp E T top = E T bottom  mgho + ½ mvo 2 = mghf + ½ mvf2  mgΔh = ½ m (v f 2 - v o 2 )  Δh = -0.5 mv o 2 /mg = 0.5v o 2 /g = (0.5)(1.79 m/s) 2 /(9.81 m/s 2 ) = 0.161 m (Use this for predicted Δh in Table 1) 11. What Δh would you get if you added 0.250 kg to the cart? Try it to make sure of your answer. (No sensor! See #8 note) It would be the same with the extra .250 kg added to the cart. The height does not depend on the change of mass. Take screenshot of motion graph of Cart with addition of 250 g mass and 500 g mass, respectively, and paste below. MOTION GRAPH OF CART WITH M C = 0.250 kg. MOTION GRAPH OF CART WITH M C = 0.500 kg Fig. 5a Motion Graph of Cart with m C = 0.250 kg Fig. 5b Motion Graph of Cart with m C = 0.500 kg Fig. 5 x-t Graph showing that height of the ramp, Δh , does not depend on the mass of Cart (m C ) 12. That doesn’t seem right. It appears that there would be no limit to how much mass could be sent through the same Δh. This would have to “cost” us somehow. Where must this extra energy be coming from? B. Experimental Δh from Δx , Δy measurements on the track and geometry. Let’s see how accurate your prediction of Δh is by measuring it. In Figure 4 you see the cart at two positions, original, xo , and final, xf . Ofcourse these would be reversed if the cart were going down the ramp. You know your ramp angle, θ . And the positions xo, and xf can be found in the apparatus data table. You can then find Δh using the geometry of the cart on the ramp shown in Fig. 7 on the right. By definition from the right-angle triangle, ∆ h =( x ¿¿ f − x o ) sin θ ¿ Figure 7 – Δh from Trigonometry

Physical Science Laboratory: PSC 151 Lab 6 VPL Lab – Dynamics, Energy and Work - mod 5 Rev 12/19/18 Mod: 2/9/19 At the bottom of the ramp, xo = 0.0740 m and at the top of the ramp, xf = 1.92 m. Using these values of x0 and xf with the angle of the ramp, w calculate Δh: ∆ h = (1.92 m – 0.074 m)sin5 o = m [ Note: make sure your calculator is set to degrees mode] 13. How close were your two values of Δh? Did you find that energy was conserved? Why or why not? The values are the same. The change in height cannot be changed by a change in mass. Energy is conserved because the energy before and after are the same. C. The relationship between speed and kinetic energy – proportional reasoning Here’s your chance to get a better understanding of the non-linear relationship between speed and kinetic energy. This is something that gives many students trouble. 1. How fast do you think you need the cart to move to reach half the maximum distance up the incline? First, we identify half the maximum distance up the incline as: x1/2f = 0.925 m and the starting position at bottom of ramp as xo = 0.074 m Next, we find Δh at half the maximum distance up the incline: Δh = (x1/2f - xo)sinθ = (0.925 m – 0.074 m)sin5o = 0.074 m Then, we use this Δh in energy conservation equation to find vo that the Cart needs to reach x1/2f: KEbottom = PEtop  ½ mvo2 = mgΔh_  vo = (2gΔh)1/2 = [2(9.81 m/s2)(0.074 m)]1/2_= 1.21 m/s = 121 cm/s____ a. Why can we not get vo the Cart needs to reach x1/2f by dividing vo to reach xf by 2? Why? You cannot get the answer by just dividing. The velocity is squared. 2. Set the initial speed of the Cart to vo to reach x1/2f obtained above. Test and see if it works. How far up the incline did the cart go? The cart went to the 0.925 m point. Fig. 8 x-t Motion Graph showing that Cart reaches half maximum distance of 1.92 m (= 0.925 m) up the incline when v o = 121 cm/s = 1.21 m/s, as calculated from energy conservation above.

Physical Science Laboratory: PSC 151 Lab 6 VPL Lab – Dynamics, Energy and Work - mod 6 Rev 12/19/18 Mod: 2/9/19 3. Aim for different positions along the track and try different speeds. What do you observe about the relationship between initial speed and different final distances on the track? The relationship between initial speed (vo) and any height h, is vo = (2g h)1/2 = [2g(xf - xo)sin5o]1/2 It is derived from energy conservation that is shown above. Experimentally, the initial speed, vo, to reach different final distances, xf, along the 5o ramp are shown below: For xf1 = 0.50 m, vo = 0.86 m/s; For xf2 = 1.00 m, vo = 1.26 m/s; For xf3 = 1.50 m, vo = 1.56 m/s; _  h at the distances will be:  h = (xf – xo)sin  Therefore, the relationship between initial speed (vo) and any height h, is vo = (2g  h)1/2 = [2g(xf - xo)sin5o]1/2, derived from energy conservation above. Note: xo = 0.074 m ______ II.Work by Friction As the cart traveled up the ramp in part IA gravity did an amount of work equal to - Wx Δ x on the cart. If the cart is released, on the way back down gravity would do an amount of work equal to + Wx Δ x . The total work for the round trip would be zero. That is, - Wx Δ x + + Wx Δ x = 0 for the round trip That’s the nature of a conservative force. When work is done in moving an object against such a force, an equal amount of work is done by the force when the object is allowed to return to its starting point. It’s as if our work is somehow “stored up” until we need it back. Of course, most of the time mechanical energy is not conserved. As you swim across a pool, the water exerts a backwards force against you. If you stop and tread water at the other side of the pool, the water will not return the favor (energy) by pushing you back to where you started! These forces that do a non-zero amount of work when an object is moved around a closed loop are called non-conservative forces . A force like a wind that acts to increase the energy of a sailboat does net positive work. Likewise, a force like friction that removes energy from a system does net negative work. In either case we can say Wnc = Δ PE + Δ KE where Wnc is the total work done by all non-conservative forces acting. Wnc can be either positive or negative. So it can increase or decrease the energy of an object. In this part of the lab we’ll look at the negative work done by friction. We’ll use a level track and observe the decrease in velocity of the cart when friction is acting. Figure 9 shows the setup. You won’t see the vectors while the car is sitting still. Figure 9 – Friction slows cart between x1 and x2 A cart traveling initially at some velocity, vo , at x = 10 cm, slows steadily as it moves from left to right. At x1 , it has a velocity, v1 . At x2 , it has slowed down to v2 . It is slowed by a kinetic friction force with some friction coefficient, µk. 1. Make sure the track is level and the brakes are on. Move the cart to the far left and set the initial velocity to somewhere in the range of 150-200cm/s.

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Physical Science Laboratory: PSC 151 Lab 6 VPL Lab – Dynamics, Energy and Work - mod 7 Rev 12/19/18 Mod: 2/9/19 a. Turn on the sensor, turn off the brakes, and press Go ->. What do you observe in the motion of the cart? The cart moves quickly to the right side of the track. The line of the graph is straight. Fig. 10 x-t Motion Graph of Cart traveling at 1.60 m/s without friction 2. Reset the sensor and cart to the initial setup above. Now change from “wheels” to “friction pad.” a. Turn on the sensor, turn off the brakes, and press Go ->. What do you observe in the motion of the cart? The cart does not go far. Fig.11 x-t Motion Graphs of Cart traveling at v o = 1.60 m/s with coefficient of kinetic friction: µ k = 0.10, 0.12, 0.14 b. How does this compare to 1a above? The first graph is straight and the line is longer. The second graph is shorter. It also starts to curve as time passes. 3. Experiment with different values of v 0 and μ k . What do you generally observe about the behavior the cart with the change to these values? Which settings allow the cart to travel farther and which make it stop sooner? From what I noticed, the lower the kinetic friction, the farther the cart will travel. As you increase the kinetic friction then the distance that they cart travels will get shorter. The cart’s distance will also depend on the velocity of the cart.

Physical Science Laboratory: PSC 151 Lab 6 VPL Lab – Dynamics, Energy and Work - mod 8 Rev 12/19/18 Mod: 2/9/19 The higher the velocity then the higher the chance of the cart moving further down the track. Fig.12 x-t Motion Graph of Cart traveling at v o = 1.90 m/s with coefficient of kinetic friction: µ k = 0.10, 0.12, 0.14 4. Friction is not all bad – in fact it is an essential part of life. Without it, the only way you would able to move would be by throwing objects away from you. Normal walking would be impossible. Let’s look at how friction holds objects in place. 5. Set the track to an angle of 12 º . The cart should fall to the bottom left. a. μ k .earlier is the coefficient of kinetic friction. It measures how much grip a surface has on a moving object. μ s conversely is the coefficient of static friction. For this part of the lab, find the value of μ s that just barely holds the cart in place. b. To do this, choose a value of μ s . and move the cart to the center of the ramp. If it falls, try a higher value. If it stays, try a lower value. Fig. 13 x-t graph showing motion Cart down 12 o incline from the middle of the ramp, at coefficient of static friction µ s = , after which the cart no longer moved c. What values of μ s did you find just barely holds the cart in place? _µ s = With kinetic friction at 0.12, The cart is held in place by static friction values of 0.22 and over.