__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 *μ _{k}* and a length

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 *v _{0}*

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* μ _{k}*

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:

E_{i}
= E_{f}

K = ½mv^{2}

E_{th} = -W_{friction}

W = F·Δs_{ǁ}

F = μ_{k}·F_{N}

5. Application of the plan of solution:

E_{i}_{ }=
E_{f}

K_{i} = E_{th} + K_{f}

½mv_{0}^{2} = E_{th} + ½mv^{2}

where E_{th} = -W_{friction}
= -[F·Δs_{ǁ}] = -[-μ_{k}·(·(2L)]

½mv_{0}^{2} = [μ_{k}·(·(2L)]
+ ½mv^{2}

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 v_{0}^{2}, 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

½mv_{0}^{2} = ½mv^{2} or v_{0}
= 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:

E_{i}_{ }=
E_{f}

K_{i} = E_{th} + K_{f}

½mv_{0}^{2} = E_{th} + ½mv^{2}

where E_{th} = -W_{friction}
= -[F·Δs_{ǁ}] = -[-μ_{k}·mg·2L]

½mv_{0}^{2} = μ_{k}·mg·2L
+ ½mv^{2}

v =

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