- Problem Solving Part 2 (physics)mr. Standring's Webware 2nd Edition
- Problem Solving Part 2 (physics)mr. Standring's Webware 2.2
- Problem Solving Part 2 (physics)mr. Standring's Webware 24
- Problem Solving Part 2 (physics)mr. Standring's Webware 2.
There are two types of problem solving activities for this class.
- Group Problem Solving (Mondays and Wednesdays)
- Problem Solving Sessions (Fridays)
Collection of thousands of eBooks please go to www. 3.2 Percentage of students agreeing with questions on the ASSIST. Physics -Problem Solving Steps To solve a physics problem, it is often useful to follow a common set of steps. You may not necessarily use all of these steps for a specific problem, and you may sometimes follow a different order of steps Read the problem. A physics guide by a physics student (not Mr. Standring) for physics students Physics? Physics – and most science subjects – can be very complicated. Describing our world is. 2.4 - 2.5: Motion Along a Line with Constant Acceleration/ Visualizing Motion Along a Line with Constant Acceleration (12) 2.6: Free Fall (12) 2: Comprehensive Problems (18) 2: Test Bank (68) Chapter 3: Motion in a Plane 3.1: Graphical Addition and Subtraction of Vectors (12) 3.2: Vector Addition and Subtraction Using Components (16).
Group Problem Solving (Mondays and Wednesdays)
These in-class problems are solved in groups and are not graded.
| SES # | TOPICS |
|---|---|
| 1 | Group problem (PDF) |
| 2 | Group problem (PDF) Group: Line of charge (PDF) Group: Uniformly charged disk (PDF) |
| 4 | Group problem: Superposition (PDF) Group problem: E from V (PDF) Group problem: Build it (PDF) |
| 7 | Group problem: Charge slab (PDF) Group problem: Charge slab (PDF) |
| 9 | Group problem: Spherical shells (PDF) |
| 10 | Partially filled capacitor (PDF) |
| 12 | Group problem: B field from coil of radius R (PDF) |
| 14 | Group problem: Non-uniform cylindrical wire (PDF) Group problem: Current sheet (PDF) |
| 17 | Group problem: Current loop (PDF) |
| 18 | Group problem: Circuit (PDF) |
| 20 | Group problem: Changing area (PDF) Group problem: Generator (PDF) |
| 21 | Group problem: Solenoid (PDF) |
| 23 | Group problem: Coaxial cable (PDF) Group problem: Circuits (PDF) |
| 28 | Group problem (PDF) |
| 30 | Superposition principle (PDF) Group problem: Plane waves (PDF) |
| 31 | Group problem: Inductor (PDF) Group problem: Capacitor (PDF) |
| 33 | Group problem: B field generation (PDF) Group problem: Energy in wave (PDF) |
Problem Solving Sessions (Fridays)
Counts toward 6% of the course grade.
| SES # | TOPICS |
|---|---|
| 3 | Coordinate systems; Gradients; Line and surface integrals (PDF - 1.4 MB) |
| 6 | Continuous charge distributions (PDF) |
| 8 | Gauss's law (PDF) |
| 11 | Capacitors (PDF) |
| 16 | Ampere's law (PDF) |
| 19 | Magnetic fields: Force and torque on a current loop (PDF) |
| 22 | Mutual inductance and transformers; Inductors (PDF) |
| 26 | RC and RL circuits (PDF) |
| 29 | Driven LRC circuits (PDF) |
| 32 | EM radiation (PDF) |
| 35 | Interference (PDF) |
Problem Solving Part 2 (physics)mr. Standring's Webware 2nd Edition
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Problem Solving Part 2 (physics)mr. Standring's Webware 2.2
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Problem Solving Part 2 (physics)mr. Standring's Webware 2.
Problem :
A 5 kg block is moved up a 30 degree incline by a force of 50 N, parallel to the incline. The coefficient of kinetic friction between the block and the incline is .25. How much work is done by the 50 N force in moving the block a distance of 10 meters? What is the total work done on the block over the same distance?
Finding the work done by the 50 N force is quite simple. Since it is applied parallel to the incline, the work done is simply W = Fx = (50)(10) = 500 J.
Finding the total work done on the block is more complex. The first step is to find the net force acting upon the block. To do so we draw a free body diagram:Because of its weight, mg, the block experiences a force down the incline of magnitude mg sin 30 = (5)(9.8)(.5) = 24.5 N. In addition, a frictional force is felt opposing the motion, and thus down the incline. Its magnitude is given by Fk = μFN = (.25)(mg cos 30) = 10.6 N. In addition, the normal force and the component of the gravitational force that is perpendicular to the incline cancel exactly. Thus the net force acting on the block is: 50 N -24.5 N -10.6 N = 14.9 N, directed up the incline. It is this net force that exerts a ìnet workî on the block. Thus the work done on the block is W = Fx = (14.9)(10) = 149 J.