Process for Problem Solving

Below is a summary of the suggested steps when solving a standard end-of-the-chapter textbook problem.  These won't apply exactly to standard AP 1 problems which are often multi-step and partly conceptual, but the ideas may still be helpful.

1. Read and understand the problem.  This may seem obvious, but it’s a good idea take a minute to thoroughly visualize what's happening.  If you can't, reread the situation until you can.

2. Draw a picture of what's happening. Even if a diagram is there, drawing-out what is happening and labeling all of the given values will reinforce the sense of what’s happening.

3. Write down a list of what is given.  Write down the variable asked for as the solution.

4. Decide what the method of solution will be.  This is often a choice of Newton's second law, conservation of energy, conservation of momentum, etc.  If there are multiple stages of motion, each stage may require its own method or model. *

5. Apply the method or methods to the stage or stages of motion.  This will take you from the given values to the answer.  If a given method or model doesn't work (say F = ma doesn't seem to get the answer), try another model (like conservation of energy).

6. Check the solution.  If the solution is numeric, you can check that the value is reasonable.  If the solution is algebraic, you can check that the units on both sides of the equation are equal (dimensional analysis).  You can also check if the result makes sense if zero is inserted for a given variable or check that it makes sense if one value increases, a second value correspondingly increases or decreases.

* If stuck in this step, try solving a simpler version of the given problem.  Let this suggest the solution to the original problem.

Below is this process applied to one of the energy problems in the practice set: A block with mass m reaches the incline shown above where both ramps are 45º above the horizontal and have a coefficient of kinetic friction μk and a length L.  Find the speed of the block at the base of the far side.

1. Visualize what is happening: The block is going to slide up the double-ramp, slowing down as it rises and generating thermal energy.  It will reach the top of the double-ramp and then slide down the far side, speeding-up as it generates even more thermal energy.

2. Draw the situation with all of the information in situ. 3. Given values:

The block has a mass m

Initial velocity is v0

The angle of each ramp above the horizontal is 45º

Each ramp hypotenuse has a length L

The coefficient of kinetic friction between the block and each ramp is μk

Solution value:

The velocity of the block after it has descended from the double ramp, v

4. Method of solution:

Conservation of energy - the total initial energy of the system will be equal to the total final energy of the system.  The total initial energy is in the form of the initial kinetic energy of the block.  The total final energy of the system is in the form of the thermal energy generated by kinetic friction and the final kinetic energy of the block.

Useful equations:

Ei = Ef

K = ½mv2

Eth = -Wfriction

W = F·Δsǁ

F = μk·FN

F = mgcos(θ) = 5. Application of the plan of solution:

Ei = Ef

Ki = Eth + Kf

½mv02 = Eth + ½mv2

where Eth = -Wfriction = -[F·Δsǁ] = -[-μk·( ·(2L)]

½mv02 = [μk·( ·(2L)] + ½mv2

v = 6. Does the solution make sense?

The equation above shows the final velocity will be less than the initial velocity, which is should be if thermal energy is generated.  The equation also shows that the final velocity will be equal to the initial velocity if μ, g, or L are zero.  This also makes sense because making any of those zero will prevent thermal energy from being generated.  Likewise, if any of those three increase, the second term under the radical will increase and a greater value will be subtracted from v02, making the final velocity lesser.  This makes sense because if μ, g, or L are increased, more thermal energy will be generated, leaving less energy for the final kinetic energy.

* What is a simpler version of this question that might lead to answering the original question?

- One simplification would be to imagine that the incline is frictionless.  Conservation of energy would then look like:

Energy before the double-ramp = Energy after the double-ramp

K = K

½mv02 = ½mv2 or v0 = v.  This may seem like a trivial result, but it’s just a warm-up for the actual question.

- Another simplification would be to imagine a level, rough surface of 2L.  Then conservation of energy gives us:

Ei = Ef

Ki = Eth + Kf

½mv02 = Eth + ½mv2

where Eth = -Wfriction = -[F·Δsǁ] = -[-μk·mg·2L]

½mv02 = μk·mg·2L + ½mv2

v = which is very similar to the original question, just absent the cos(θ) term in the normal force equation.