TPS for Problem Solving Dr. Edward Prather
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1 TPS for Problem Solving Dr. Edward Prather University of Arizona Center for Astronomy Education (CAE)
2
3 Insights from the Univ. of Arizona AAU STEM reform effort in Physics Reformed Class Two 50 minute lectures per week Focused on introducing concepts using active engagement instructional strategies and on collaborative group problem solving Minimal derivations of equations Each student also attends a 50 minute recitation sections per week Led by graduate TA with assistance from undergraduate peer instructors Students work on collaborative tutorials, which promote reasoning abilities and problem solving skills Instructor experienced in astronomy and physics education research, but teaching PHYS 141 for the first time Traditional Class Three 50 minute lectures per week Focused on introducing concepts and on instructor-led modeling of problem solving Many derivations of equations Instructor experienced in teaching PHYS 141 and widely regarded by faculty and students as an excellent lecturer
4 Chapter 6: Work and Kinetic Energy F θ d A constant force displacement. F is applied to an object. The object has a straight-line d The work W by the force on the object is given by W = F d = Fd cos( θ) SI unit of work: Joules (J) dot product aka scalar product (see Section 1.10) Force and displacement are vectors, but work is a scalar.
5 F d When a force (or component of a force) points in the same direction as the displacement, the work done by that force is positive (W > 0). d F When a force (or component of a force) points in the opposite direction as the displacement, the work done by that force is negative (W < 0). d F When a force is perpendicular to the displacement, the work done by that force is zero (W = 0).
6 In the cases below (1-5), identical particles experience the same displacement. The forces shown acting on the particles all have the same magnitude. Rank each case based on the work done on the particle, from most negative to most positive. Case 1 Case 2 Case 3 d F d A) 1, 3, 5, 2, 4 ) 5, 3, 4, 1, 2 C) 5, 3, 1, 4, 2 D) 4, 2, 5, 3, 1 E) None of the above. d F F Case 4 Case 5 d F d F
7 When many forces act on an object, we can calculate the work done on that object by each of those forces. The sum of all of these works is called the net work (W net ). A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net.
8 Which of the following is the correct free-body diagram for the box? A f I N I N I C f I N I f I W E W E W E D f I N I E N I f I W E W E
9 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E Is the work done by the normal force (W N ) on the box positive, negative, or zero? A) positive ) negative C) zero W N = ( N I )( d)cos(90 o ) = 0 J
10 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E Is the work done by the friction force (W f ) on the box positive, negative, or zero? A) positive ) negative C) zero
11 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E Is the work done by the weight force (W W ) on the box positive, negative, or zero? A) positive ) negative C) zero
12 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E Work done on box by friction force: W f = ( f I )( d)cos(180 o ) = ( f I )( d) f I is given by which of the following?
13 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E Work done on box by friction force: W f = ( f I )( d)cos(180 o ) = ( f I )( d) f I is given by which of the following? A) µ k mg cos ( ) ) µ k mg sin ( ) C) µ k mg cos (60 o ) D) µ k mg E) More than one of the above.
14 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E The work done by the weight force (W W ) is given by which of the following?
15 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E The work done by the weight force (W W ) is given by which of the following? A) mg d ) mg d sin (60 o ) C) mg d cos ( ) D) mg d cos(60 o ) E) More than one of the above.
16 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E What is the net work (W net ) done on the box?
17 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E What is the net work (W net ) done on the box? A) 34.5 J ) 29.4 J C) 24.3 J D) 5.1 J E) None of the above.
18 A 3.0 kg box starts from rest and slides for 2.0 m down a incline. The coefficient of static friction between the box and the incline is µ k = 0.1. Find W net. coordinate axes y free-body diagram f I N I x W E W N = ( N I )( d)cos(90 o ) = 0 J ( ) ( ) d W f = ( f I )( d)cos(180 o ) = ( µ k ) mgcos ( )= 5.1J W W = ( W E )( d)cos(60 o ) = ( mg) ( d)cos(60 o )= 29.4 J W net = W N W f W W = 0 J 5.1J 29.4 J = 24.3J
19 COPUS data from UA Calc-Physics Reformed Course
20 Exam 1 Average (%) Reformed (N = 206) Traditional (N = 234) 0.00 Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Entire Exam Exam Item
21 20.00 Exam Reformed - Trad. Scores (%) Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Entire Exam Exam Item
22 Exam Percentage of students Reformed (N = 206) Traditional (N = 234) Grade on Exam 1 (points)
23 Exam 2 Average (%) Reformed (N = 206) Traditional (N = 226) 0.00 Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Entire Exam Exam Item
24 20.00 Exam Reformed-Trad. Scores (%) Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Entire Exam Exam Item
25 Exam Percentage of students Reformed (N = 206) Traditional (N = 226) Grade on Exam 2 (points)
26 Exam 3 Average (%) Reformed (N = 203) Traditional (N = 230) 0.00 Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Entire Exam Exam Item
27 20.00 Exam Reformed - Trad. Scores (%) Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Entire Exam Exam Item
28 Exam Percentage of students Reformed (N = 203) Traditional (N = 230) Grade on Exam 3 (points)
29 Final Exam Average (%) Reformed (N = 217) Traditional (N = 258) 0.00 Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Item 7 Item 8 Item 9 Entire Exam Exam Item
30 25.00 Final Exam Reformed - Trad. Scores (%) Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Item 7 Item 8 Item 9 Entire Exam Exam Item
31 35.00 Final Exam Percentage of students Reformed (N = 217) Traditional (N = 258) Grade on Final Exam (points)
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