Group Problem Week 9

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Physics

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Apr 3, 2024

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I Physics 131 Group Problems: Week 9 Leaming Goals After working through this worksheet, students will be able to: 1. Draw a coordinate system either aligned with the force or with the displacement vector; use it to find the parallel component of the other; and use the parallel component to find the work done by the force. 2. Graphically determine the work done by a force using a force vs. position graph. 3. Determine the net work done on an object or system by adding up the work done by each individual force acting on it. 4. Use the net work done on an object or system to determine its change in kinetic energy. Problem 1: Calculating work using force and displacement components To find the work done on an object by a force, we need to define a new coordinate system that aligns with either the force or the displacement vectors. This is just like what we did when finding the torque a force exerts on an object; the difference is that we're now concerned with the parallel component of either the force or the displacement. A train moves 20 km along a track to the north-east. As it moves, the wind blows against the side of the train, exerting a 150 kN force at an angle of 60° from the displacement. ,s• w: ... ,s O ' 00 " ,.1 A. Compute the work done on the train using the parallel component of the force: W = F 11 lctl, where dis the displacement and F 11 is the component ' · : of the force parallel to the displacement. ,. 1. On the diagram at right, sketch a coordinate system which has one coordinate axis parallel to the displacement, and one perpendicular to the displacement. / l.: 2. Use your i:oordinate system to find the parallel component of the force. F, , :. IF I CA>l~ =- \So \L ' "' o,~ . (oo 1 \!-i, 3. How much work does the wind do on the train? v-> f II I?-\ ; 7 Soo o -1- ,z..,o coo= \ -c; • ,c, .. J B. Compute the work done on the train using the parallel component of the displacement: W = d 11 IFI, where.Fis the force and d 11 is the component of the displacement parallel to the force. 1. On the diagram at right, sketch a coordinate system which has one coordinate axis parallel to the force, and one perpendicular to the fore~. F 2. Use your coordinate system to find the parallel component of the displacement. ~ ., ,.,[ a\ c...o~ -:.1.,,0 c"~\oO ::. \Oo 0 o~ 3. How much work does the wind do on the train? w==- O-t1 Ir/= (oOO' \"St:> (J'>b -:. \,S· 10" 1 1

Problem 2: Work due to a single force Three ice skaters are skating across an outdoor ice rink while a wind exerts a drag force on them. The skaters each move the same displacement to the right, and the magnitude of the drag force is the same. The direction of the drag force is different in each case, as shown in the diagrams at right. Case I ,i ~J . VJ-:;~.~ A. Rank the three skaters according to the magnitude of the work done by the wind on them, from largest to smallest. If the work is zero for any skater, state so explicitly. J ,i ,2 F;~ ' = f ~e,o\e I > -2. > J •~ c:i L"'" ~" ~~\' . . . . c.,,..,,_~\•-..J ' B. Is the kmetlc energy of any of the skaters mcreasmg? If so, which ones? Case 3 F;-s! LI J C. Suppose the wind blew on a fourth skater such that the skater's kinetic energy decreased. In the space below, sketch two different possible diagrams for this skater, showing the direction of their displacement and the direction of the wind force on them. (There are an infinite number of possible diagrams that would all work, each with different angles and magnitl,ldes of the drag force. You should draw two of them.) .,. C, ,,, _t ;.1 ;, t "' ' D. Compute the work done by the wind force on the skater for case 2, if the magnitude of the wind force lfff is 12 N, the angle that the force vector makes with the horizontal direction is 40°, and the magnitude of the displacement vector is 2 m. 2

Problem 3: N et work done on a single object by multiple forces In this problem we will compute the net work done by several different forces on an object and use the work-energy relation W = t:.K to compute the change in the kinetic energy of the object. Consider a person moving a box across a floor by exerting a force Fpull, with a magnitude of 80 N at an Fpull angle of 40° from the negative x-axis. The mass of the box is 10 kg, and the floor exerts a friction force on the box with a magnitude of 45 N. The box is dragged 3 m to th.e left. A. Draw a point free-body diagram of the box. ~J~F•qs, .,. B. Since the displacement is horizontal, it will be easie's't to use a coordinate system aligned C. with the displacement and find the work usirig" the parallel component of each force. For each force, first find its parallel component (including the sign) and then find the work it does. 1. The gravitational force: Fu= 0 W= O--s 2. The normal force: Fu= 0 W= O-S 3. The pull force: Fi1 = '60cos ~v a W= 0\. ' 'b r-- ' • :s = ,~')_,,'1--S 4. The friction force: Fi1 = -~~t-1 W= r-11 · ~ -: -\~6'S What is the net work done on the object? 6vJ : v,) I -I i,J '2. t W 1' v-J L\ :: 0-+ 0 +- 1"(1 .Gi - l?. s -- 4 .~ 1 D. What is the kinetic energy of the box after it has been dragged for a distance of 3 m, assuming that it started at rest? vJ = I<... 6 =- ~i - \c , Lt i. = - == "'i · 1

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Problem 4: Work done by gravity and change in gravitational potential energy In problem 3, the box was moving in a horizontal direction, and therefore no work was do: ~y gravity. Here, we will analyze a situation where the force of gravity has some component a points along the directio~ of the displacement, and therefore there is non-zero work done by gravity on the system of interest. Consider a box of mass 10 kg, initially at rest, which slides down a frictionless incline, as shown in the figure at right. The incline has a length of 5 m and makes _an angle of 20° with the horizontal. For now, consider the system of only the box, so that the force of gravity does work on the system. · '5 A. Draw a free-body diagram of the box. Next to your diagram, show the direction of the displacement of the box. . B. Find the work done by each force on your free-body diagram. 1. Work due to the normal force: w~ o 2. Work due to the gravitational force: F,,::. t.o~ 1 0• : \_,..)-= s \)>~ 10 .'::' \_ µ ~(.., J,. c. What is the kinetic energy of the box when it reaches the bottom of the incline? t; =O· : )}l~t-1t-:: b.l'.:. .__ ~i: - ~: , h ::. \J'l ... (A D. Calculate the speed of the block at the bottom of the slope. \ r 1r 1.., ..... .,, .r;.'l- 4 \J Now suppose there is friction between the block and the ramp with µ = 0.2 . E. What is the work done by the force of friction on the box sliding down the ramp? (Hint: Recall that you'll need to find the normal force, which will balance the component of the gravitational f~rce that is pointing perpendicular to the ramp .) ~\'...= ~N ::: P""J e,o~ g w~_\"j.a 4 \

F. Use the work done by friction and your previous results to determine the speed of the block at the bottom of the slope in this case. Is it bigger or smaller than what you got without friction? Does your answer make sense? Wt\ e., \-;::; 6 y:. .: \C~ \l~ ;:: v-1 ('\' +- \N f :. -\ I'<)\}~ '2... -I'- S l'(\t1. \\tr -1-V\M i,.1 ; -\-., 011.\ - kicA-' '" " l \flt, l' · A- 'o'I. ) Cw , \ "' -q ;) G. If the friction force is large enough, the block will slide at constant speed down the ramp, and thus l!..K = 0. 1. For this to be true,' how much work must be done by the friction force? \Nf\ eA-::. () W t, +-\)c-=o \N \:: _ vJ~ . 2. Use your answer and the length of the ramp to determine the magnitude of the friction force for the block to move at constant speed. \e"~~°" o~ ' """ i' :. 5 {V\ \{'J1--=- -t!L a ::. F (( = tJ 3. Check your answer by using Newton's 2° d law to find the friction force such that . there will be no acceleration down the ramp. r- Gt c.os (3 .,.. F It _.,,. c::, fc 1 (I> \ { 'HI- e ", Finally, let's consider the system of the block plu~ the earth, so th~t gravity is no longer an external force. Recall that the gravitational potential energy is given by UG = mgy, where y is the vertical position of the box (pick whatever origin you want). H. What is the change in the potential energy (l!..U) of the box as it slides from the top to the bottom of the incline? L>M = M) L:1 ') " 0 6".'.)= (s,..,., \ Si'I\L<) I. · How does your answer compare to the work done by gravity that you found in part B?

Problem S: Work as the area under a force-vs-position curve The work done by a force can also be detennined graphically, by finding the area under a graph of the force vs position, analogous to how we could find the displacement as the area under the velocity vs. time graph. A. The simplest case to find the work for is when the force and displacement are parallel. Consider a case where F = 4 N pointing to the right, and the object moves to the right from x = 0 m to x = 5 m. A plot of the force vs. position for this case is shown at right. 1. Use W = Fu Id I to find the work that this force does on the object. F 5m 2. Shade the area under the graph from x = 0 m to x = 5 m, and find this area. B. Suppose the force still pointed to the right, but the ~bject moved from x = 5 m to x = 0 m. Would this change the area under the curve? Would this change the work done? f\t"-. S"""-(_ (\•,~~ \(l. tt;~ \- "n , I "-"'~y,\\- · , \(t, I ' I 1J, -'LO -s ,\ ' . ·. . I ', ' . ' ' I \ \ ' " \' .A 1 J J 1 ,h..1 l As your answers above show, this graphical method gives us the absolute value of the work. To detennine the sign, we need to compare the directions of the force and displacement. C. Sketch a force vs. position graph for a force which does negative work on an object . which is moving to the right. Shade the area corresponding to the work done. D. The box in problem 3 is pulled for 15 m across a carpeted room, with the given friction force of 45 N. The next room has a hardwood floor, which drops the friction to 15 N for the next 15 mas shown in the graph. On the graph, shade the area corresponding to the work done by the friction force, and then use the graph to find the total work done by friction. Should your answer be positive or negative? (\t,. , n.\'\ ¼ _, · 1l.- \\ 'fV\ oJ IJ i,._ X -)Om ~I = I) t\.-t- v'\tG -)" \_'\,t 'I. • Jr ett'•"' 6 -15 m F \ !SN "

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r Problem 6: Work done against a spring force and spring potential energy We can use this graphical idea to find the work done to stretch or compress a spring (which is also equal to the spring potential' energy). Consider a horizontal spring that is attached to a block at one end and tethered to a wall at the other end. If we pull (or push) on the block, such that the block moves at a constant speed, then the net force on the block must be zero. Thus lftpuul = IFsPI = kl~xl. If we consider the system of the block and spring, the only external force that can do work is the force of our hand on the block (the spring force is internal to the system). A plot of the force we exert on this spring is shown at right. A. Suppose we stretch the spring from x = 0 cm to x = S cm. Shade the area on the graph corresponding to the work we've done on the system. Then compute the work we've done . w ,, , \ { ~~\ · Io) == 0 1- 1: S --'S B. For a point particle, or a simple system like a box, external work can only transfer energy into kinetic energy. What form of energy is our work transferring energy into? C. The spring potential energy is given by U 5 = i kx 2 for a spring stretched along the x- , axis. Given the area you sketched and your answer above, explain why this makes sense. (Hint: At some horizontal position x, the height of the graph is equal to the spring force, which is kx). w= A= 1 b · "1 ;:; t • f"/, , l r.-,): D. The spring is now compressed from x = 0 cm to x = -5 cm. Sketch the area on the graph corresponding to the work we do in this case, in a different color or shading pattern. Then compute the work we've done. How does the sign of the work in this case compare to when we stretched the spring? 7

l E. One of the experiments that helped determine the mechanical properties of duplex DNA involved attaching a bead to each end of the strand and stretching it. A force vs. extension graph of this experiment is shown at right. The y-axis shows the force (in pN) and IIO r--------------, 70 60 50 z the x-axis plots the stretch per base-pair (in nm) .e«> Consider two different extensions of DNA: from 0.30 to j )0 0.34 nm, and from 0.42 to 0.46 nm. Which extension required more work? (You do not need to compute the 10 -ONA °'""m, t1c hlf-t rq_•o,,, r-- - · --i work done. Just answer this question qualitatively.) 0 - ~..:i+, -, .... ,.,. ......... ,..,..,.+-..:,.+,+-,--.~~ 0. 26 0.54 On the graph, shade the areas corresponding to each amount of work. 8

Group Problem Week 9

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