Physics_114_Practice_Exam_3_Solutions

PHYSICS 114 PRACTICE EXAM 3 Name (Print): PID: Honor Pledge: On my honor, I have neither given nor received unauthorized aid on this examination. Signature: Do not open the exam until told to do so. You will have 50 minutes to complete the examination. NO CELL PHONES, TEXT MSG, etc. ALLOWED AT ANY TIME. Before the exam begins: • Print and sign your name, and write your student PID number in the spaces above. During the exam • When the exam begins, print your name on each page. Do this first when you are told to open your exam. • If you are confused about a question, raise your hand and ask for an explanation. • If you cannot do one part of a problem, move on to the next part. • This is a closed book examination. All equations and constants are provided. • You may use a calculator, but not a computer, or other Internet-connected devices (smart-phones, iPads, etc.). • Please write neatly and legibly. Do not use scratch paper; it will be ignored. • Show your work in enough detail so that the grader can follow your reasoning and your method of solution. Circle your answers, and state units if appropriate. End of exam: • Out of respect to other students, please remain seated for the last 10 minutes of the exam. At the end of the exam, please remain seated until all exams have been collected. PHYS 114

Physics 114, Practice Exam 3 2 1. [8 pts] A 70-kg student decides to bungee jump off a bridge that is 65 m above a river. The bungee cord attached to his ankles acts like a spring with an average spring constant of 900 N/m. How long should the bungee cord be (when unstretched) so that he goes all the way to the surface of the water, but does not go into the water? Show your work. The stretched length of the bungee cord must be 65 m. This must be equal to the unstretched length plus the amount by which the cord has stretched: If we consider the student, the bungee cord, and the Earth to be a system, then there are no external forces acting on the system. At the start of the jump, all of the system’s energy is gravitational potential energy. When the cord is fully stretched, the student will be at rest. At this instant, he will be at rest and all of the energy will be elastic potential energy. We can use this to solve for the amount by which the cord is stretched: If we know the amount by which the cord is stretched, then we can determine its unstretched length:

Physics 114, Practice Exam 3 4 3. [8 pts] On a sled, you slide without friction down a small, ice-covered hill. If you start from rest at the top of the hill, your speed at the bottom is 7.50 m/s. In search of cheap and transient thrills, you decide to get a running start. If your speed is 1.50 m/s at the top, what is your speed at the bottom of the hill? Show your work. If the system is defined as including the sled, person, and Earth, then there are no external forces that do work on the system. When you start from rest at the top, all of the initial energy in the system is in the form of gravitational potential energy; by the time you reach the bottom, it has been converted into kinetic energy. We can use this information to figure out the height of the hill: When you get a running start, the system initially has both kinetic and gravitational potential energy. All of this becomes kinetic energy at the bottom of the hill. We can use this to find the final speed:

Physics 114, Practice Exam 3 5 4. [5 pts] The graphs below (A-C) each show how the magnitude of a force applied to an object changes as the object moves a certain distance. Assume that the object in each case is moving in the same direction as the force points and assume all three objects are identical. Each object only experiences a single force. Rank the graphs based on the change in kinetic energy experienced by the objects, from greatest to least. Explain your reasoning. The area under the curve corresponds to the work done by the force, which equals the change in the kinetic energy. Both graph B and C have the same area under the curve. Graph A has the smallest area under the curve. Therefore, the correct ranking is B = C > A. 0 1 2 3 4 0 1 2 3 4 5 6 7 8 force (N) distance (m) 0 1 2 3 4 0 1 2 3 4 5 6 7 8 force (N) distance (m) 0 1 2 3 4 0 1 2 3 4 5 6 7 8 force (N) distance (m) A B C

Physics 114, Practice Exam 3 7 b. [6 pts] Determine the magnitude of the force from that the latissimus dorsi and pectoralis major muscles exert on a single arm. Show your work. At right is the extended free-body diagram for a single arm. We don’t know the magnitude or direction of the force from the shoulder joint, so I just represented it by an x- and a y-component. However, if I put the pivot point at the shoulder joint, then I will only have non-zero torques from the force from the muscles and the force from the ring. Since the gymnast is in static equilibrium, the net torque must be zero, so we can use this to figure out the unknown force from the muscles: c. [5 pts] Determine the magnitude of the force from the shoulder joint on the arm. Show your work. Since the gymnast is in static equilibrium, the net force acting on the arm must be zero. We can use this to figure out the x- and the y-components of the force from the shoulder joint: Now use the Pythagorean theorem to find the magnitude of the force from the shoulder joint: d. [3 pts] Determine the angle the force from the shoulder joint makes with respect to the horizontal. Clearly indicate whether it points above or below the horizontal. Show your work. The force from the shoulder joint will point above the horizontal at an angle of F 1A F MA F x F y

Physics 114, Practice Exam 3 8 6. A box slides down an inclined plane that makes an angle of 30 o with respect to the horizontal. There is friction between the box and the inclined plane and there is a spring at the bottom of the inclined plane. The figures below show the position of the box at three times: • t 1 : The box is sliding down the inclined plane, toward the spring. • t 2 : the box has collided with the spring, which is fully compressed. • t 3 : the box has rebounded off of the spring and is traveling back up the inclined plane. Note that the speed at time t 3 is less than the speed at time t 1 ( v 3 < v 1 ). The box is also higher up the incline at t 1 than at t 3 . a. [4 pts] Define the system to include the box, the spring, and Earth. From t 1 until t 3 , state whether the following quantities are positive (+), negative (-), or zero (0). No explanation is necessary to receive full credit. • W net,ext (net work done by forces external to the system): – • Δ K (change in the system’s kinetic energy): – • Δ U g (change in the system’s gravitational potential energy): – • Δ U s (change in the system’s spring potential energy): 0 b. [5 pts] Define the system to include the box, the spring, and Earth. From t 2 until t 3 , is the absolute value of the change in the system’s gravitational potential energy ( Δ U g ) greater than, less than, or equal to the absolute value of the change in the system’s spring potential energy ( Δ U s )? Explain your reasoning. With this definition of the system, the friction force will be doing work on the system, while the gravitational and spring forces will cause a change in gravitational and spring potential energy, respectively. The work-energy theorem therefore becomes 𝑊𝑊 𝑛𝑛𝑛𝑛𝑛𝑛 , 𝑛𝑛𝑒𝑒𝑛𝑛 = ∆𝐸𝐸 𝑊𝑊 𝑓𝑓 = ∆𝐾𝐾 + ∆𝑈𝑈 𝑔𝑔 + ∆𝑈𝑈 𝑠𝑠 The change in kinetic energy is positive, since the box speeds up. The work done by friction is negative since the friction force points in the opposite direction of the box’s displacement. The box’s height is increasing, so the change in gravitational potential energy is positive. The change in the spring potential energy is negative because the spring is decompressing. Looking at the above equation, the left side (W f ) is negative, which means the right side must also be negative. Since there is only one negative term on the right side ( ΔU s ), it must have larger absolute value than the sum of the two positive terms ( ΔK + ΔU g ). Therefore, the absolute value of ΔU s is greater than the absolute value of ΔU g . 𝑣𝑣 ⃗ 1 t 1 𝑣𝑣 ⃗ 2 = 0 m/s t 2 𝑣𝑣 ⃗ 3 t 3 30 o 30 o 30 o

Physics 114, Practice Exam 3 10 𝑥𝑥 = � 𝑚𝑚𝑣𝑣 𝑖𝑖 2 + 2 𝑚𝑚𝑚𝑚 (0.08 m) − 2( 𝜇𝜇 𝑘𝑘 )( 𝑚𝑚𝑚𝑚 )cos(30 𝑜𝑜 )(0.16 m) 𝑘𝑘 = � (2.5 kg)(0.45 m/s) 2 + 2(2.5 kg)(9.8 m/s 2 )(0.08 m) − 2(0.15)(2.5 kg)(9.8 m/s 2 )cos(30 𝑜𝑜 )(0.16 m) 565.6 N/m = 0.08 m Note: You should the same answer regardless of your choice of system. If you include the incline in the system, then instead of having a negative work done by friction, you would have a positive change in thermal energy term (with the same absolute value as the work done by friction). Similarly, if you left Earth or the spring out of the system, then instead of having change in gravitational and spring potential energies, respectively, you would have work done by these forces; these work terms would have the same absolute value, but opposite sign, as their associated change in potential energy terms.

Physics 114, Practice Exam 3 11 7. [11 pts] The ancient Egyptians raised giant monuments, called obelisks, which were carved out of a single piece of stone. The figure at right is a simplified diagram representing one method the Egyptians may have used to move an obelisk into a standing position. The obelisk has a mass of 413,000 kg and a height of 32 m. The bottom the obelisk rests on the edge of a “turning groove,” while the top touches a wall at a point 28 m above the top of the turning groove. You may assume there is no friction between the obelisk and wall, although the wall does exert a 50,000-N normal force on the obelisk. The obelisk makes an angle of 75 o with respect to the horizontal. A rope attached 14 m above the obelisk’s center of mass is used to pull the obelisk into its final, upright position. What is the maximum tension force in the rope while the obelisk is in static equilibrium? Show your work. The figure at right shows the extended free-body diagram for the obelisk. It includes the normal force from the wall (N WO ), the tension force from the rope (T RO ), the weight force from Earth (W EO ), and the x- and y-components (F x and F y , respectively) of the unknown contact force from the turning groove. Since we know nothing about this contact force, we should place our pivot point there so that it has a torque of 0 N-m. Since the obelisk is at a 75 o angle with respect to the horizontal, N WO must make an angle of 75 o with respect to its r-vector, and W EO must make an angle of 15 o with respect to its r-vector. The magnitude of the r-vector for W EO is 12 m, for T RO it is 26 m, and for N WO it is (28 m)/sin(75 o ) = 29 m. We can apply the condition of net torque = 0 N-m to solve for the unknown tension force: 𝜏𝜏 𝑇𝑇 + 𝜏𝜏 𝑁𝑁 − 𝜏𝜏 𝑊𝑊 = 0 ( 𝑇𝑇 𝑅𝑅𝑅𝑅 )(26 m)sin(35 𝑜𝑜 ) + ( 𝑁𝑁 𝑊𝑊𝑅𝑅 )(29 m)sin(75 𝑜𝑜 ) − ( 𝑊𝑊 𝐸𝐸𝑅𝑅 )(12 m)sin(15 𝑜𝑜 ) = 0 ( 𝑇𝑇 𝑅𝑅𝑅𝑅 )(26 m)sin(35 𝑜𝑜 ) = ( 𝑊𝑊 𝐸𝐸𝑅𝑅 )(12 m)sin(15 𝑜𝑜 ) − ( 𝑁𝑁 𝑊𝑊𝑅𝑅 )(29 m)sin(75 𝑜𝑜 ) 𝑇𝑇 𝑅𝑅𝑅𝑅 = ( 𝑊𝑊 𝐸𝐸𝑅𝑅 )(12 m)sin(15 𝑜𝑜 ) − ( 𝑁𝑁 𝑊𝑊𝑅𝑅 )(29 m)sin(75 𝑜𝑜 ) (26 m)sin(35 𝑜𝑜 ) 𝑇𝑇 𝑅𝑅𝑅𝑅 = (413,000 kg)(9.8 m/s 2 )(12 m)sin(15 𝑜𝑜 ) − (50,000 N)(29 m)sin(75 𝑜𝑜 ) (26 m)sin(35 𝑜𝑜 ) 𝑇𝑇 𝑅𝑅𝑅𝑅 = 749,000 N wall center of mass 75 o turning groove 35 o 12 m 14 m 28 m 50,000 N T RO N WO W EO F y F x

Physics 114, Practice Exam 3 13 b. [5 pts] In reality, there is friction between Thomas and the wooden train track, so Thomas only gets part way up the bridge. At the instant his speed is 0 m/s on the bridge, he is 2.75 cm above the floor. From the instant Thomas begins moving until he stops part way up the bridge, what is the total amount of work that friction does on the Thomas-Earth system? Show your work. Since the system includes Earth and Thomas, the work-energy theorem becomes 𝑊𝑊 𝑛𝑛𝑛𝑛𝑛𝑛 , 𝑛𝑛𝑒𝑒𝑛𝑛 = ∆𝐸𝐸 W 𝑓𝑓 = ∆𝐾𝐾 + ∆𝑈𝑈 𝑔𝑔 W 𝑓𝑓 = 𝐾𝐾 𝑓𝑓 − 𝐾𝐾 𝑖𝑖 + 𝑈𝑈 𝑔𝑔 , 𝑓𝑓 − 𝑈𝑈 𝑔𝑔 , 𝑖𝑖 Both the initial and final kinetic energies are 0 J, so this becomes W 𝑓𝑓 = 𝑈𝑈 𝑔𝑔 , 𝑓𝑓 − 𝑈𝑈 𝑔𝑔 , 𝑖𝑖 W 𝑓𝑓 = 𝑚𝑚𝑚𝑚�𝑦𝑦 𝑓𝑓 − 𝑦𝑦 𝑖𝑖 � = (0.057 kg)(9.8 m/s 2 )(0.0275 m − 0.055 m) = − 0.015 J

Physics 114, Practice Exam 3 14 9. The dinosaur Diplodocus had a 6.5-m-long neck and its head and neck had a total weight of 13,400 N. The head and neck were supported by a ligament that exerted a tension force 0.42 m from the joint where the neck attached to the body (note: the joint is marked by the red dot on the figure below). The magnitude of the tension force the ligament exerted has been estimated as 404,000 N when the ligament is assumed to have made an angle of 10 o with respect to the neck (note: the center of the neck is represented by the dashed red line on the figure below). For both part a and part b, imagine that Diplodocus is not moving its head and neck. a. [7 pts] Determine the distance x from the joint (red dot) of the center of mass of the head and neck. Show your work. Since the neck and head are at rest, we can apply the conditions of static equilibrium. If we put the pivot point at the joint (red dot), then the condition of net torque = 0 N-m becomes (404,000 N)(0.42 m)sin(10 o ) - (13,400 N)(x) = 0 𝑥𝑥 = (404,000 N)(0.42 m) sin(10 𝑜𝑜 ) (13,400 N) = 2.2 m b. [6 pts] What is the magnitude of the force the joint exerted on the neck? Show your work. The force from the joint has an unknown x-component (F x ) and an unknown y-component (F y ). We can determine these components by applying Newton’s first law to the x- and y-components individually. Start with the x-components: F x – (404,000 N)cos(10 o ) = 0 F x = (404,000 N)cos(10 o ) = 398,000 N Now apply Newton’s first law to the y-components: F y + (404,000 N)sin(10 o ) – (13,400 N) = 0 F y = (13,400 N) – (404,000 N)sin(10 o ) = -56,800 N It is okay that the result for F y is negative; that just means that F y points down (in the same direction as the 13,400 N force) instead of up as I assumed. The magnitude of the force from the joint can now be calculated by inserting the x- and y- components of the force into the Pythagorean theorem: 𝐹𝐹 = � (398,000 N) 2 + (56,800 N) 2 = 402,000 N head 13,400 N 0.42 m x 404,000 N

Physics 114, Practice Exam 3 16 c. [4 pts] Determine the magnitude of the frictional force that the ground exerts on the ladder. Show your work. We can look at the forces acting on the ladder in the x-direction to determine the frictional force that the ground exerts on the ladder:

Physics 114, Practice Exam 3 17 g = 9.8 m/s 2 k B = R/N A = 1.38 x 10 -23 J/K 1 in. = 2.54 cm R = 8.31 J/mol-K 1 ft. = 0.305 m N A = 6.02 x 10 23 mol -1 1 mi. = 1609 m 1 atm = 1.01 x 10 5 Pa �𝑆𝑆 ⃗ � = 𝑘𝑘 | 𝑥𝑥 ⃗ | r C circle π 2 =