Physics 114 General Physics I Lecture 28 PDF

12/2/24 Physics 114 General Physics I: For Students of the Life Sciences Lecture 28...

12/2/24 Physics 114 General Physics I: For Students of the Life Sciences Lecture 28 Prof. Yue Qiu Please check in with the Email: [email protected] UNC Check-In app. Office: 178 Phillips Hall © 2024 All rights reserved. 1 Announcement  Our Final Exam is on Dec 12th, 8:00-11:00 AM.  The Final Exam will be held in three rooms Last names beginning A-I will be in Genome Science G100 Last names beginning J-Sherwood will be in Hamilton Hall 100 Last names beginning Shoffner-Z will be in Chapman 211 There will not be a seating chart.  The make-up is on Dec 13th, 8:00-11:00 AM. You can only take the make-up if your request form is approved by the Advising Office.  If this make-up time still does not work for you, please e-mail Prof. Zhang ([email protected]) ASAP to arrange an alternative option.  More information about the Final Exam location and format can be found in the Canvas announcement from Prof. Zhang.  3 Today’s Class  Review topics covered  Sample problems © 2024 All rights reserved. 12/2/24  4 Topics Covered – 1  Scaling  How does changing the dimension of something affect its other properties?  Mathematical models  Log-Log and Semi-Log  Kinematics: 𝑥(𝑡), 𝑣(𝑡), 𝑎(𝑡)  How are they related to each other?  Solving problems graphically and with equations given constant acceleration. © 2024 All rights reserved. 12/2/24  5 Topics Covered – 2  Newton’s laws of motion, 𝐹Ԧ = 𝑚𝑎Ԧ  Free-Body Diagrams, motion in an elevator  Impulse and momentum  How do these relate to force?  What changes the motion of an object?  Stress and strain  Young’s modulus, stress/strain curves  Resilience  Torque, COM, equilibrium, and applications  Static equilibrium: 𝐹Ԧ𝑛𝑒𝑡 = 0, 𝜏Ԧ𝑛𝑒𝑡 = 0 © 2024 All rights reserved. 12/2/24  6 Topics Covered – 3  Work, kinetic energy, potential energy  Energy conservation equation  Internal vs. external work  Oscillations, damped oscillations  Determining phase  Damping constant, time constant © 2024 All rights reserved. 12/2/24  7 Topics Covered – 4  Thermodynamics, ideal gas law  Conserving energy – the first law  Interpreting PV-diagrams  2nd law of thermodynamics  Diffusion  Diffusion scaling, rms distance © 2024 All rights reserved. 12/2/24  8 Equation Sheet © 2024 All rights reserved. 12/2/24  9 Class Response Question A crossing guard holds a stop sign at arm’s length, as shown. Her arm is horizontal, and we assume that the deltoid muscle is the only muscle supporting her arm. The weight of the upper arm is Wu = 18 N, the weight of the lower arm is Wl = 11 N, the weight of her hand is Wh = 4.0 N, and the weight of the stop sign is Ws = 8.9 N. A force of magnitude fd is exerted by the deltoid on the humerus and the shoulder joint exerts a force on the humerus whose x- and y-components are given by fx and fy, respectively. Calculate the magnitudes of fd, fx, and fy. © 2024 All rights reserved. 12/2/24  10 Class Response Question Identify which of the following equations are correct according to Newton’s laws. ( ) f x = fd cos 18o fy = f sin (18 ) d o ( ) f y = Wu + Wl + Wh + Ws + f d sin 18o fy = W + W + W + W - f sin (18 ) u l h s d o f y = Wu + Wl + Wh + Ws © 2024 All rights reserved. 12/2/24  11 Class Response Question From the condition that the net force = 0 N, we have two equations…but 3 unknowns! f x = fd cos 18o ( ) ( ) f y = Wu + Wl + Wh + Ws - f d sin 18o © 2024 All rights reserved. 12/2/24  12 Poll Question 28.1 Now we need to apply the condition of net torque = 0 N-m in order to find a 3rd equation. Where should we put the pivot point in order to get an equation that only includes one of the unknown forces? A) where the deltoid attaches to the humerus B) where the shoulder joint attaches to the humerus C) where the hand holds the stop sign D) where the weight of the upper arm acts E) where the weight of the lower arm acts © 2024 All rights reserved. 12/2/24  13 Poll Question 28.2 Which of the following equations is correct (using the sign convention)? ( A) - f d cos 18o ( )) (14 cm) + (W ) (18 cm) u ( )( ) ( )( ) ( )( + Wl 42 cm + Wh 65 cm + Ws 65 cm = 0 ) B) - ( f sin (18 )) (14 cm) + (W ) (18 cm ) d o u + (W ) ( 42 cm ) + (W ) (65 cm ) + (W ) ( 65 cm) = 0 l h s C) ( f cos (18 )) (14 cm) - (W ) (18 cm) d o u - (W ) ( 42 cm) - (W ) ( 65 cm) - (W ) ( 65 cm) = 0 l h s D) ( f sin (18 )) (14 cm) - (W ) (18 cm) d o u - (W ) ( 42 cm) - (W ) ( 65 cm) - (W ) ( 65 cm) = 0 l h s E) none of the above © 2024 All rights reserved. 12/2/24  14 Class Response Question From the condition that the net force = 0 N, we have two equations: f x = fd cos 18o ( ) f y = Wu + Wl + Wh + Ws - f d sin 18o ( ) From the condition that the net torque = 0 N-m, we have one equation: ( f sin (18 )) (14 cm) - (W ) (18 cm) d o u ( )( ) ( )( - Wl 42 cm - Wh 65 cm - Ws 65 cm = 0) ( )( ) © 2024 All rights reserved. 12/2/24  15 Class Response Question Solution: fd = ( W ) (18 cm ) + (W ) ( 42 cm) + (W ) ( 65 cm) + (W ) (65 cm ) u l h s (sin (18 )) (14 cm) o fd = (18 N ) (18 cm) + (11 N ) ( 42 cm ) + ( 4.0 N ) ( 65 cm) + (8.9 N ) ( 65 cm) = 375 N (sin (18 )) (14 cm) o ( ) ( ) ( ) f x = fd cos 18o = 375 N cos 18o = 357 N f y = Wu + Wl + Wh + Ws - f d sin 18o ( ) ( ) ( ) ( ) ( ) ( f y = 18 N + 11 N + 4.0 N + 8.9 N - 375 N sin 18o = -74 N ) ( ) © 2024 All rights reserved. 12/2/24  16 Class Response Question A box of mass m2 is on a 30° ramp. The coefficient of kinetic friction between the box and the ramp is μk. A string of negligible mass attaches to one end of this block, loops around a frictionless pulley, and connects to the top of a second box, of mass m1 , as shown below. Assume that m1 has an acceleration that points downward. What is the magnitude of m1’s acceleration? (Write an expression for it in terms of the given variables.) m1 30o © 2024 All rights reserved. 12/2/24  17 Class Response Question 2nd law equation for 𝑚1 : TS1 – WE1 = m1a1 2nd law equation for 𝑚2 in y-direction: NR2 – WE2 cos(30o) = 0 2nd law equation for 𝑚2 in x-direction: - fR2 - WE2 sin(30o) + TS2 = m2a2 We can reduce the number of unknowns in our equations: The two masses share the same magnitude of acceleration. Call this magnitude 𝑎 (𝑎 = −𝑎1 = 𝑎2 ). If the string is massless, then the tension is the same everywhere on that string. Call this tension magnitude T ( T = TS1 = TS2 ). We were given the coefficient of kinetic friction, so 𝑓𝑅2 = 𝜇𝑘 𝑁𝑅2 © 2024 All rights reserved. 12/2/24  18 Class Response Question 2nd law equation for 𝑚1 : TS1 – WE1 = m1a1 2nd law equation for 𝑚2 in y-direction: NR2 – WE2 cos(30o) = 0 2nd law equation for 𝑚2 in x-direction: - fR2 - WE2 sin(30o) + TS2 = m2a2 − 𝑎1= 𝑎2 = 𝑎 Replace variables TS1 = TS2 = T 𝑊𝐸1 = 𝑚1 𝑔 𝑓𝑅2 = 𝜇𝑘 𝑁𝑅2 𝑊𝐸2 = 𝑚2𝑔 Let’s re-write our equations: 𝑇 − 𝑚1 𝑔 = − 𝑚1 𝑎 𝑁𝑅2 − 𝑚2 𝑔 cos 30° = 0 −𝜇𝑘 𝑁𝑅2 − 𝑚2 𝑔 sin 30° + 𝑇 = 𝑚2 𝑎 © 2024 All rights reserved. 12/2/24  19 Class Response Question 1 𝑇 − 𝑚1 𝑔 = −𝑚1 𝑎 Remember that we are 2 𝑁𝑅2 − 𝑚2 𝑔 cos 30° = 0 given 𝑚1 , 𝑚2 , 𝜇𝑘 , and 𝑔. And we are solving for 𝑎. 3 −𝜇𝑘 𝑁𝑅2 − 𝑚2 𝑔 sin 30° + 𝑇 = 𝑚2 𝑎 © 2024 All rights reserved. 12/2/24  20 Class Response Question Remember that 𝑎 is the magnitude of acceleration for both 𝑚1 and 𝑚2 (𝑎 > 0), so 𝑚1 𝑔 > 𝜇𝑘 𝑚2 𝑔 cos 30° + 𝑚2𝑔 sin 30° must be true. © 2024 All rights reserved. 12/2/24  21 Work done ON the gas ▪ The work done on the gas is equal to the area under the pV curve. ▪ This work done on the gas will be positive when the gas is compressed (volume gets smaller) and negative when the gas expands (volume gets bigger). p W p p W W V V V ▪ The work done BY the gas has the opposite sign. © 2024 All rights reserved. 12/2/24  22 Poll Question 28.3 For this isochoric process: A. Heat and work on system are positive. B. Heat is negative and work on system is positive. C. Heat is positive, work on system is zero. D. Heat and work on system are negative. E. Heat is zero, work on system is positive. © 2024 All rights reserved. 12/2/24  23 Poll Question 28.4 For this isothermal process A. Heat and work on system are positive. B. Heat is negative and work on system is positive. C. Heat is positive, work on system is zero. D. Heat and work on system are negative. E. Heat is zero, work on system is positive. © 2024 All rights reserved. 12/2/24  24 Poll Question 28.4 This is an isothermal process, which means that the temperature of the system doesn’t change. If the temperature doesn’t change the thermal energy doesn’t change. Therefore DT = 0 ® DETh = 0 DETh = Won + Q = 0 In this case, since the work done on the system is positive ( the gas is being compressed and the work done on the gas is equal to the area under the PV diagram curve), the Q must be negative. Heat flowed out of the system. DETh = Won +Q = 0 ® if Won > 0 ® Q < 0 © 2024 All rights reserved. 12/2/24  25 Poll Question 28.5 The figure at right shows two different processes taking an ideal gas from state i to state f. Is the temperature change during process A larger than, smaller than, or equal to the change during process B? A. Larger than B. Smaller than C. Equal to © 2024 All rights reserved. 12/2/24  26 Poll Question 28.6 The figure at right shows two different processes taking an ideal gas from state i to state f. Is the heat exchanged during process A greater than, less than or equal to the heat exchanged during process B? A. Greater than B. Less than C. Equal to © 2024 All rights reserved. 12/2/24  27 Poll Question 28.7 1 mole of hydrogen gas (H2) i follows the process shown in the PV diagram at right. The thermal energy in the final state of the gas is _____ …the initial state’s. A. Greater than f B. Less than C. Equal to © 2024 All rights reserved. 12/2/24  28 Poll Question 28.8 1 mole of hydrogen gas (H2) i follows the processes shown in the PV diagram at right. Is the magnitude of the heat exchanged during this process greater than, less than, or equal to the magnitude of f the work done on the gas? A. Greater than B. Less than C. Equal to © 2024 All rights reserved. 12/2/24  29 Poll Question 28.9 What is the phase constant for the oscillation shown in the graph below? 0.1 s A. 1.9 rad B. ⎯ 0.63 rad C. ⎯ 1.9 rad D. 0.1 rad E. ⎯ 1.3 rad © 2024 All rights reserved. 12/2/24  30 Poll Question 28.10 For a certain population of bacteria, the number of bacteria n is related to time t (in hours) by the following function: () log n = 0.22t + 3.7 Which of the following correctly rewrites this function in the form y = nakx? ( ) A) n = 103.7 100.22t B) n = ( e ) e 3.7 0.22t C) n = (10 ) e 3.7 0.22t D) n = ( e )10 3.7 0.22t E) none of the above © 2024 All rights reserved. 12/2/24  31 Poll Question 28.11 As a bungee jumper falls from a bridge, when is her speed the greatest? A) When the bungee cord first starts to stretch beyond its equilibrium point B) When the bungee cord stretches enough to equal her weight C) When the bungee cord stretches the maximum amount © 2024 All rights reserved. 12/2/24