Test 3 solution Problem 1: Short Answer Questions / Multiple Choice a. > 1 b. > 4 c. > 9 d. > 8 e. > 9 Problem : Estimation Problem (a GOAL Approach student solution) While this is a good GOAL approach to the problem, there are three points that need attention. The 1.5 m stretch of the net goes all the way to the ground, no room for error. In the numeric substitution, there are no units & the x term (1.5 m) is not squared. There are two many digits in the final answer, one or two significant digits would suffice.
Problem (another more compact approach student solution) This is a nice elegant approach. Note the safety factor of 0.5 m. Only one small nit to pick, the answer also has too many significant digits > k 78,000 N/m would have been sufficient. Problem 3: Essay Question Since this is an essay question some writing is preferred Good student solution, but more elaboration on why cart B accelerates at a higher rate than cart A is needed.
Problem 4: Student Solution This solution hits all the key points but is light on explanations. In the absence of gravity (negligible force), the pie and the astronaut stick together > totally inelastic collision There are no external forces so the momentum of the astronaut-pie system is conserved Since this is an inelastic collision, some mechanical energy (in this case KE) is lost due to Work by non-conservative forces. Corrections: Answer has the wrong units Newtons, not Joules. No unit check Again, too many significant digits in the answer > Wnc - 800 J Thermal Energy generated - Wnc 800 J
Problem 5: Problem 5 was on the nd practice test (with solution) and was assigned as WebAssign homework (student solution). Need to pick a coordinate system and use it consistently. Since the Normal force points up and the Weight force points down, they cannot both be positive. Assuming up is the +y direction, the mg term here should be negative. This will yield a final answer for part A: N mv r T C + mg.5 x 10 4 N Almost a good solution. Note the use of the free-body diagrams with good labels and the explicit use of Newton s nd law. For part b, the solution correctly shows that the Normal force > 0 for the case where the velocity is a maximum. corrections Again the final answer has too many digits. The length of the arrows representing the force vectors in the free-body diagram do not reflect the relative magnitudes of the two force vectors. The answer for part a is incorrect. There is a sign error in the derivation. The normal force and the weight force cannot both be positive. If up is the +y direction, the weight force is (-).
GROUP PROBLEM (Note that I am using GOAL to solve this problem as I am typing it, calculations done in excel from numeric calculation shown here. No scratch paper was used.) Gather: Truck m truck 0 metric tons x 1000 kg / m t 0,000 kg Point 0: top of the h 0 50 m v 0 100km/h * 1000m/km * 1h/3600s 7.8 m/s is inclined 0 degrees µ truck-road? Point 1: bottom of the bottom of the v 1 175km/h * 1000m/km * 1h/3600 sec 48.6 m /s h 1 0 m Point : top of the v 0 m/s h? is inclined 30 degrees µ truck-road 1.5 h 50 m x Ramp x h? θ θ Assume that the truck does not slide during this problem Want to find: The coefficient of static friction between the truck and the and the distance the truck travels ukp the before it comes to a stop Guesstimates: µ rubber-concrete ~ 1 > this should be µ for the truck on the h > The truck loses mechanical energy going down the and up the, so the h < h 0 and we know that with the speed the truck has at the bottom that h > 0, so 0 < h < h 0 f Organize: Free body diagram Truck going down the truck N truck Truck going up the N truck f truck (static) W earth truck W earth truck ( ) θ W earth truck ( ) What is happening? Down the Truck goes down the speeding up Gravitational Potential energy is converted in KE and thermal energy (Work by friction) Define coordinate system so that + x points down the and +y point in the direction of the N force Up the Truck slows down as it goes up the Kinetic Energy is converted into Gravitational PE and thermal energy (Work by friction) Define coordinate system so that + x points up the and +y point in the direction of the N force
Group Problem page Physics principles at work: Conservation of energy Work Energy Theorem Friction force Part 1. Find the coefficient of friction between the truck and. Since we don t need to know how long it took the truck to get down the, use conservation of energy to find the work done by non-conservative forces (friction). We can find initial PE, final PE, initial KE, and final KE. Can derive conservation of energy from work energy theorem Use the definition of work and the relationship between friction and normal forces to find µ Will use geometry to get distance truck travels down the Part. How high up is the truck when it stops? Since we know beginning speed and final speed and don t need to know time, use Work-Energy Theorem The net work done is the work done by gravity and the work done by friction Use geometry to relate displacement of truck to h Analyze: General Derivation of conservation of energy equation W net KE W c + W ext + W nc KE ( PE) + W ext + W nc KE note: W c PE W ext + W nc KE + PE (KE f KE i ) + (PE f PE i ) PE i + KE i + W ext + W nc PE f + KE f (General Conservation of Energy Equation) Down the PE 0 + KE 0 + W ext + W nc PE 1 + KE 1 but PE 1 0 J (at base of, h 1 0 m). Also If my system includes the and the and we neglect air resistance, then W ext 0 J since there are no external forces PE 0 + KE 0 + W nc KE 1 (m t gh 0 ) + (½ m t v 0 ) + (f road truck x cos α) (½ m t v 1 ) recall that f µ N and cos α is the angle between the frictional force and the displacement, α 180 m t gh 0 + ½ m t v 0 + (µn road truck ) x ( 1) ½ m t v 1 recall that N road truck m t g cos θ & cos α 1 m t gh 0 + ½ m t v 0 µ (m t g cos θ ) x ½ m t v 1 divide both sides by m t gh 0 + ½ v 0 µ g cos θ x ½ v 1 Solve for µ: µ ½v0 + gh0 g x cosθ 1 We know everything on the right hand side (RHS) except x But sin θ h 0 / x so x h 0 / sin θ, substituting back in our equation for µ ½v + gh ½v + gh sinθ + gh 0 0 1 0 0 1 0 0 1 µ tan h gh0 cosθ gh 0 0 g cosθ sinθ ½v θ
Group Problem page 3 ( 48.6 m / s) (0.5)(7.8m / s) + (9.8m / s)(50m) (0.5) tan 0 (9.8m / s)(50m) µ 0.5 note that µ is a unitless quantity Up the PE 1 + KE 1 + W ext + W nc PE + KE but KE 0 J (object at rest) and PE 1 0 J (at base of, h 1 0 m). Also if my system includes the and the and we neglect air resistance, then W ext 0 J since there are no external forces. KE 1 + W nc PE (½ m t v 1 ) + (f road truck x cos α) (m t gh ) recall that f µ N and cos α is the angle between the frictional force and the displacement, α 180 ½ m t v 1 + (µn road truck ) x ( 1) m t gh recall that N road truck m t g cos θ ½ m t v 1 µ(m t g cos θ ) x m t gh divide both sides by m t ½ v 1 µ g cos θ x gh We know everything in this equation except h and x But sin θ h / x so h x sin θ, substituting back in our conservation of energy equation v 1 g cosθ x g µ x sinθ µ g x cosθ g x sin v 1 + θ > x( µ g cosθ + g sinθ ) v1 x g( µ cosθ + sinθ ) (9.8m / s (48.6m / s) )(1.5* cos(30 ) + sin(30 )) 67 m v 1 Learn step: