Problem Solving Part 2 (physics)mr. Standring's Webware 2

There are two types of problem solving activities for this class.

Problem Solving Part 2 (physics)mr. Standring
  • Group Problem Solving (Mondays and Wednesdays)
  • Problem Solving Sessions (Fridays)

Springer is part of Springer Science+Business Media ( Dedicated to my parents. For solving the problems can be located in the beginning of each chapter. Chapters 2 and 3 focus on quantum physics. Chapter 2 is basically concerned. Kinematic equations relate the variables of motion to one another. Each equation contains four variables. The variables include acceleration (a), time (t), displacement (d), final velocity (vf), and initial velocity (vi). If values of three variables are known, then the others can be calculated using the equations. This page demonstrates the process with 20 sample problems. At one level, problem solving is just that, solving problems. Presented with a problem you try to solve it. If you have seen the problem before and you already know its solution, you can solve the problem by recall. Solving physics problems is not very di erent from solving any kind of problem. Problem Solving (CPS) is a useful tool for teaching physics through problem solving. Using Cooperative Problem Solving. These four chapters describe the foundation of cooperative problem solving and provide detailed information about how to implement CPS for maximum effectiveness. This information.


Group Problem Solving (Mondays and Wednesdays)

These in-class problems are solved in groups and are not graded.

Group problems solving.
1Group problem (PDF)

Group problem (PDF)

Group: Line of charge (PDF)

Group: Uniformly charged disk (PDF)


Group problem: Superposition (PDF)

Group problem: E from V (PDF)

Group problem: Build it (PDF)


Group problem: Charge slab (PDF)

Group problem: Charge slab (PDF)

9Group problem: Spherical shells (PDF)
10Partially filled capacitor (PDF)
12Group problem: B field from coil of radius R (PDF)

Group problem: Non-uniform cylindrical wire (PDF)

Group problem: Current sheet (PDF)

17Group problem: Current loop (PDF)
18Group problem: Circuit (PDF)

Group problem: Changing area (PDF)

Group problem: Generator (PDF)

21Group problem: Solenoid (PDF)

Group problem: Coaxial cable (PDF)

Group problem: Circuits (PDF)

28Group problem (PDF)

Superposition principle (PDF)

Group problem: Plane waves (PDF)


Group problem: Inductor (PDF)

Group problem: Capacitor (PDF)


Group problem: B field generation (PDF)

Group problem: Energy in wave (PDF)

Problem Solving Sessions (Fridays)

Counts toward 6% of the course grade.

Problem solving sessions.
3Coordinate systems; Gradients; Line and surface integrals (PDF - 1.4 MB)
6Continuous charge distributions (PDF)
8Gauss's law (PDF)
11Capacitors (PDF)
16Ampere's law (PDF)
19Magnetic fields: Force and torque on a current loop (PDF)
22Mutual inductance and transformers; Inductors (PDF)
26RC and RL circuits (PDF)
29Driven LRC circuits (PDF)
32EM radiation (PDF)
35Interference (PDF)


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Problem :

Problem Solving Part 2 (physics)mr. Standring's Webware 24

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?

Problem Solving Part 2 (physics)mr. Standring's Webware 2.

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.

Problem Solving Part 2 (physics)mr. Standring's Webware 2.2

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.