Homework answers / question archive / PHYSICS SAMPLE LAB WRITE-UP Title - Newton’s 2nd Law Objective In this experiment we will attempt to confirm the validity of Newton’s 2nd Law by analyzing the motion of two objects (glider and hanging mass) on a horizontal air-track

PHYSICS SAMPLE LAB WRITE-UP Title - Newton’s 2nd Law Objective In this experiment we will attempt to confirm the validity of Newton’s 2nd Law by analyzing the motion of two objects (glider and hanging mass) on a horizontal air-track

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PHYSICS SAMPLE LAB WRITE-UP Title - Newton’s 2nd Law Objective In this experiment we will attempt to confirm the validity of Newton’s 2nd Law by analyzing the motion of two objects (glider and hanging mass) on a horizontal air-track. First, we will calculate the theoretical acceleration by applying Newton’s 2nd Law (Fnet = MA), neglecting friction, to the glider and hanging mass. Next, we will calculate the experimental acceleration of the glider by applying the kinematic equations of motion as it moves between two markers (photogates) on the track. We will then compare the experimental acceleration to the theoretical acceleration. Theory a) Acceleration using Newton’s 2nd Law Apparatus Setup V1 Photogates V2 glider M1 d +X Airtrack M2 +Y hanging mass Free-Body Diagram N T T M1 M1g M2 M2g Apply Newton’s 2nd Law to mass M1 and M2. Mass ‘M1’ ΣFx = T = M1a Mass ‘M2’ 1 ΣFY = M2g - T = M2a Adding both equations gives: M2g = M1a + M2a atheo = M2g/(M1 + M2) b) Acceleration using Kinematic Equations Using the kinematic equation V22 = V12 + 2a ( x − x0 ) we will calculate the experimental acceleration of the glider as it moves between the two photogates. We will take the origin of our coordinate system at the first photogate. d = distance between photogates V1 = (s/t1) velocity of the glider through photogate 1 V2 = (s /t2) velocity of the glider through photogate 2 s = diameter of small flag on glider t1 = time for small flag to go trough photogate 1 t2 = time for small flag to go trough photogate 2 a exp = V22 − V12 2d Apparatus Refer to theory section for apparatus setup One air track(#21), blower(#2), blower hose and power supply One digital photogate(#2C) and one accessory photogate(#2A) One glider(#1B) One flat accessory box(#22A) String Electronic pan balance(#1) Vernier Calypers (#12c) Procedure 1. 2. 3. 4. 5. Measure the mass of the glider and hanging mass. Setup the air track and blower as indicated by instructor. Measure the distance between photogates. Measure the diameter of the small flag on glider with vernier calipers. Release glider 10 cm away from photogate 1 and record times trough both photogates. 6. Repeat step (5) four more times. 2 Data M1= 4750 g M2=50.00 g g = 9.80 m/s2 d = 60.65 cm s = 1.01 cm Run # t1 t2 1 2 3 4 5 0.039 0.043 0.044 0.041 0.038 0.023 0.024 0.023 0.023 0.032 V1 (cm/s) 25.5 23.0 22.5 24.5 26.0 V2 (cm/s) d (cm) 43.0 41.5 42.5 42.5 43.5 60.65 60.65 60.65 60.65 60.65 aexp (cm/s2) 9.91 9.86 10.7 9.97 10.1 Calculations Theoretical Acceleration: atheo = M2g/(M1 + M2) = 50.00 g*980 cm/s2/(4750g + 50.00 g) atheo = 10.2 cm/s2 Experimental Acceleration: a exp = V22 − V12 = (43.5 cm/s)2 - (26.0 cm/s)2 /(2*60.65 cm) (sample calculation Run #5) 2d aexp = (9.91 +9.86+10.7+9.97+10.1)/5 = 10.1 cm/s2 (average experimental acceleration) % error = exp− theo % error = ? theo 10.1−10.2 10.2 × 100 ? X 100 = 0.98 % 3 Conclusion 1. The theoretical acceleration using Newton’s 2nd Law was 10.21 cm/s2 and the average experiment acceleration using the kinematic equations was 10.10 cm/s2. The percent error between experiment and theory was only 1%. Although the percent error was small, there were still systematic and random errors present. 2. Based on the relative small % error of 0.98% we can conclude that the objective of confirming Newton’s 2nd Law was accomplished. 3. In measuring the velocity of the gliders through the photogates we used the average velocity instead of the instantaneous velocity. This resulted in the average velocity always being smaller than the instantaneous velocity. This will V 2 − V12 then cause a exp = 2 to be consistently smaller than atheo which resulted in a 2d systematic error. A second systematic error was that in applying Newton’s 2nd Law to derive atheo of the glider we neglected the frictional force. The resulting equation should have been atheo = (M2g – fk)/(M1 + M2). Neglecting friction on the atheo equation should result in atheo being consistently larger than aexp. The data shows this to be true with the exception of one data point. 4. In addition to the random errors involved due to the uncertainty of the measuring devices, other random errors involved in the experiment include: a) Not releasing the glider from same initial point every run. b) Trying to balance the air track. c) Having the hanging mass M2 swinging when releasing M1 from rest. All these random errors contributed to the uncertainty in the final results for the accelerations. These random errors also contributed to the 0.98% error in the final results. 4 NEWTON’S LAWS OF MOTION Newton’s Laws are the foundation of Classical (Newtonian) Mechanics. They were published by Isaac Newton in 1687 along with the law of gravitation in the Principia. They have far reaching applications/implications in every branch of science. Nonetheless, these laws have their limitations. They need to be modified when applied to objects moving near the speed of light or when applied to microscopic particles. As we will see Newton’s Laws are very simple to state but are NOT so simple to apply. Only with practice does one become better at learning to apply Newton’s Laws. Newton’s Laws are empirical laws, deduced from experiment; they cannot be derived from anything more fundamental! Critical to understanding Newton’s Laws of Motion is the concept of FORCE. What is a force? Let’s begin with the 4 fundamental forces of nature: 1. 2. 3. 4. Gravitational Force Electromagnetic Force Strong Nuclear Force Weak Interaction All forces that we encounter in nature fall into one of these types of forces. What are some common forces that you’ve encountered and experienced? 1. 2. 3. 4. 5. 6. Gravity (your weight) Electricity Pushes Pulls Physical contact Collisions In general, any time two or more objects interact, there is a force involved. FORCE – The interaction between two or more objects or the interaction between an object(s) and the surrounding environment. Also very important in understanding Newton’s Laws of Motion are the terms net (resultant) force and external force. Resultant Force – The vector sum of the forces acting on a body. F = F1 + F2 + F3 + ....... (Resultant Force) Ex. F2 F1 F = F1 + F2 + F3 m F3 1 External Force – Forces exerted on a body by other bodies in the environment (surroundings) m1 m2 F2(ext) F1(ext) m F3(ext) m3 FIRST LAW A body at rest remains at rest and a body in uniform motion (constant velocity) remains in uniform motion unless it is acted on by a non-zero net (resultant) external force. v=0 v=0 M M F2 ?F=0 “body remains at rest” F1 ?F = F1 + F2 = (-F2) + F2 = 0 “body remains at rest” v = constant v = constant M F2 ?F=0 “body continues at constant v” F1 M F1 = -F2 M F1 F1 = -F2 ?F = F1 + F2 = (-F2) + F2 = 0 “body continues at constant v” F2 ?F = F1 + F2?0 “body cannot be at rest or move at constant v” F1 ? F2 2 SECOND LAW The net (resultant) external force on a body is equal to the product of the body’s mass and acceleration. F ma Newton’s 2nd Law In component form, 1. Fx max Fy ma y Fz maz F is the net (resultant) external force on a body. 2. The quantity ma is not a force. The quantity ma is a vector equal in magnitude and direction to the net external force acting on a body. 3. Note that the 1st Law is just a special case of his 2nd Law when a M F 0 and thus a 0. F1 F F2 M a= F/M F a= F/M M a = (F1 + F2)/M THIRD LAW If two bodies interact, the force that body 1 exerts on body 2 is equal and opposite to the force that body 2 exerts on body 1. 1. The two forces are called an action-reaction pair and always act on different bodies. The two forces are due to the interaction between two different objects. 2. Action-reaction forces have the same magnitude but opposite direction. 3. N3L is true if the bodies are at rest or in motion. 3 Ex. Newton’s 3rd Law Consider a block on a table-top as shown below: M a) What are the forces acting on the block? b) What is the reaction force to each force acting on the block? c) What are the action-reaction pairs? FT on B FB on T a) FB on E b) M EARTH FE on B c) FT on B = - FB on T FE on B = - FB on E Units of Force MKS (SI) cgs British Eng. F ma 1 Newton(N) = 1 kg • m/s2 1 dyne(d) = 1 g • cm/s2 1 pound (lb) = 1 slug • ft/s2 Mass and Weight Inertia – A measure of the tendency of resistance of an object to changes in its initial state of motion. Mass - The property of an object that specifies how much inertia it has. It is an intrinsic property of the object. Weight - A measure of the gravitational force exerted on an object. 4 Consider an object in free-fall near the earth’s surface: m +y ay= g ?Fy= w Earth Applying N2L to the object: ΣFy = may w mg Weight of a Body Ex. m = 100 kg W E = mgE = 100kg•9.8 m/s2 = 980N (on earth) W m = mgm = 100kg•1.6 m/s2 = 160N (on moon) WE/ Wm ≈ 6 or W m = (1/6)W E Using Newton’s Laws When applying Newton’s Laws to a system we are only interested in the external force acting on the system. A system can be a single particle or object or several particles or objects. Once you define a system, the system boundary divides the universe into the system and the environment outside the system External Force – a force exerted on a system by bodies outside the system. 5 m4 system boundary m1 m2 F42 F31 F32 m3 F53 m5 F53 = external force F42 = external force F31 = internal force F32 = internal force A diagram of the external forces acting on a system is called a Free-Body Diagram. Consider a block of mass M being pulled to the right by a rope on a frictionless surface. n v M M T mg Free-Body Diagram The force N is called a normal force. It is a contact force perpendicular to the surface. STEPS IN USING NEWTON’S LAWS 1. 2. 3. 4. 5. 6. Define your system and then draw a sketch of system. Draw the Free-Body Diagram of system. Insert a coordinate system. Find the x and y-components of each external force on system. Apply N2L along the x and y-axis separately. Solve for the required quantity. USING NEWTON’S 1ST LAW (SYSTEMS IN EQUILIBRIUM) Def: A body at rest or moving with constant velocity is said to be in equilibrium. F F ma 0 0 (since a = 0) Fx 0 Fy 0 Condition for Body in Equilibrium Fz 0 6 < Edit San Jose Microsoft Word -LWEGIAN 1 ACK, Drewer WIL TIOSU W TU. 2. Two photogates 3. glider 4. accessory box 5. string 6. pan balance ? a mig {{ x =*=ma mao {Fyz = mag-f=m29+ Mzg=(M + Mzja THEORY oyi glider M ?? — theo! M?9 mitmz (T T Emig 10 M2 hanging mass - To mig umza 1. Apply N2L separately to both M and M2. 2. Derive the theoretical expression for the acceleration of the blocks (system). Conclusion 1. The theoretical acceleration using Newton's 2nd Law was 10.21 cm/s2 and the average experiment acceleration using the kinematic equations was 10.10 cm/s2. The percent error between experiment and theory was only 1%. Although the percent error was small, there were still systematic and random errors present. 2. Based on the relative small % error of 0.98% we can conclude that the objective of confirming Newton's 2nd Law was accomplished. then cause dexp = 3. In measuring the velocity of the gliders through the photogates we used the average velocity instead of the instantaneous velocity. This resulted in the average velocity always being smaller than the instantaneous velocity. This will V2 - V, to be consistently smaller than atheo which resulted in a 2d systematic error. A second systematic error was that in applying Newton's 2nd Law to derive atheo of the glider we neglected the frictional force. The resulting equation should have been atheo = (M2g – fk)/(M1 + M2). Neglecting friction on the atheo equation should result in Atheo being consistently larger than aexp. The data shows this to be true with the exception of one data point. 4. In addition to the random errors involved due to the uncertainty of the measuring devices, other random errors involved in the experiment include: a) Not releasing the glider from same initial point every run. b) Trying to balance the air track. c) Having the hanging mass M2 swinging when releasing M1 from rest. All these random errors contributed to the uncertainty in the final results for the accelerations. These random errors also contributed to the 0.98% error in the final results.