Solving slope problems.

Let’s start simply with a frictionless inclined plane and build from there, since the steps are largely the same for any block on an inclined plane like the one pictured here:

Getting the forces, probably the hardest part
The first part of solving an inclined plane problem is figuring out all of the forces acting on the block. We’ll start by looking at gravity, which has 2 components. The first is the component of gravity that tries to slide the block down the plane, called the sliding force (F=mg sin x). The second is the component of gravity forcing the block against the plane, called the y component of the block’s weight (F=mg cos x). Since I never could remember which component was sin and which was cos, I’ll give you an easy way to just figure it out. Think about if the angle was 0 (i.e., the plane was flat). Then there’s no force pushing the block down the plane. Now, which trig function gives us 0 with an angle of 0? It’s sine. Likewise, the force pushing down is the full force of gravity; which trig function gives us 1 with an angle of 0? That would be cosine.

Friction, it justs rubs you the wrong way.
Ok, so now you’ve got the force sliding the block down the plane and the force pushing it against the plane; what next? Well, you need to consider, is there friction? If not, you’re done; lucky you. The net force is just the sliding force; you can use F = ma to get the acceleration, mass, or whatever they’re asking for.

So if you’re not lucky, they put in friction. The next question is whether the force is enough to overcome static friction. Maybe they’ll tell you that the block is in fact moving or has just started to move. Cool, in that case, static friction has been breached; you get to figure out dynamic friction, and you can skip to the next section. Of course, these problems are rarely that simple, and you’ll probably have to work through static friction. The good news is that it’s not that much more work. Static friction seems complex given that the formula is F < or = uN, where u is the coefficient of static friction and N is the normal force. The normal force is the force that the incline plane pushes up against the block. (It is very convenient to keep the block from sinking into the inclined plane that the normal force does this.  ) Anyway, N is equal in magnitude to the force of gravity pushing the block down into the plane, i.e., F = mg cos x, the y-component to the weight of the block. However, N points in the opposite direction of the y component of the weight.

So what’s the deal with the inequality anyway? What static friction tries to do is generate exactly enough force to resist motion. So if we had a 10N force sliding the block down the plane and static friction could generate anything up to 100N to resist this motion, it would actually generate a 10N force in the opposite direction and your block will just sit there. However, you just need to calculate the maximum force of static friction by using the formula F = uN. (Oh, and if you used signs and got a negative number, make the result positive, or you’ll probably confuse yourself.) So ask yourself at this point, "Is the force sliding this block down the plane greater than the maximum force of static friction?" If you answered no, then the prof threw you a break and the block isn’t going anywhere. Otherwise, we can now forget that we ever figured out static friction and go to the next section.

Dynamic (Kinetic) Friction
We have good news: no inequalities are needed for dynamic friction, and the formula is exactly the same:

Fdynamic friction = uN

Ok, if your block is moving, and friction is present, you need to subtract the friction force opposing the motion from the x component of the gravitational force. This gives you the net force. If we call the x component of the gravitational force the sliding force, we would get the following equation for the net force:

Fnet = Fsliding - Fdynamic friction

Oh, if somehow you got a negative number at this point, you definitely screwed something up, since negative forces in this context mean that the block is sliding up the plane for no real reason. (Of course, if you had a couple of rocket engines strapped to the block, maybe it really IS sliding up the plane, but somehow I doubt that.)

SPEED vs VELOCITY

As far as speed is concerned, remember that speed is a scalar (just a number) and velocity is a vector (a number and a direction). You could come up with a situation where there is no change in speed but a change in velocity. That situation would be a particle moving in a circle without speeding up or slowing down (i.e., uniform circular motion). The velocity will be changing because the particle is not moving in a straight line. Even though the numerical value of the velocity isn’t changing, the direction part is changing, and therefore the velocity is also changing.

ACCELERATION, VELOCITY, AND SPEED

Acceleration is defined as a change in velocity over some time.

So if the net force is not zero, then every single time without fail no matter what, the acceleration is not zero and the velocity is changing.

If the net force is constant, the acceleartion is constant and the velocity is changing at a constant rate.

Say you have a constant acceleration of 10 m/s^2 down.

Every second, you add 10 m/s down to your previous velocity vector. If you start out going 5 m/s down, in 1 second you’ll be going 15 m/s down, in two you’ll be going 25 m/s down, etc….

If you have non-zero acceleration, but it is NOT constant, your velocity vector will change, but it will change different amounts every second. Eg, start out at 5 m/s down, next second your at 40 m/s down, the next second you’re at 2 m/s to the left.

Again, if the acceleration is constant, and it is not zero, there IS a change in velocity. Always, every single time, no matter what. It could be a change in the direction, the speed, or both. On the MCAT, acceleration problems will generally be constant non-zero acceleration. All of those problems will involve a change in velocity. Everyone, everytime, every year, every MCAT administration. Now repeat after me:

"Acceleration IS a change in velocity. If there is acceleration, the velocity is changing."

HINT:

Try to stop thinking about speed. Speed doesn’t really make sense. If you’re going 50 mph, do you care what direction you are going? Yes, of course you do. Doing your thing around town do you ever care about speed without direction? No.

Speed is an abstract silly thing that is used when bragging about the Porsche they give you if you get a 45 on the MCAT. In physics and real life, you always deal with direction at the same time, so you want to think about velocity.

Coulomb’s Law describes what happens when two charges are placed close to one another. The magnitude of the force between the two charges, called the electrostatic force, depends on two variables: the magnitudes of the two charges, and the distance between the two charges. Whether the force is attractive or repulsive depends on the signs of the two charges: like charges will repel, and unlike charges will attract. Mathematically, two charges that repel will have a positive value for the electrostatic force, while two charges that attract will have a negative value for the electrostatic force.

Dependence on the Magnitudes of the Charges:
The electrostatic force is directly dependent on the magnitude of each of the two charges. If one charge’s magnitude is doubled, the electrostatic force between them will be doubled. If the second charge’s magnitude is doubled, again, this will double the force between them. If the magnitudes of both charges are doubled, the force between them will be quadrupled.

F q1
F q2

Dependence on the Distance Between the Charges:
The electrostatic force is indirectly dependent on the square of the distance between the two charges. This is why Coulomb’s Law is known as an inverse square law, like Newton’s Law of Universal Gravitation. If the distance between the two charges is doubled, the force is cut by one-fourth. If the distance between the two charges is halved, the force between them is quadrupled.

F 1/(r^2)

Putting all of these variables together with a permittivity constant gives us Coulomb’s Law:

F = k(q1)(q2)/(r^2)

You should understand and memorize Coulomb’s Law, but you do NOT need to memorize the value of k for the MCAT.

It may be counterintuitive, but in a pendulum, the tension in the string during an oscillation is larger than mg. Here is an example to explain why this is so. Imagine for a moment that you are looking at a weight on a string that’s bouncing, and imagine that you are also looking at a pendulum swinging, but that you’re looking at it from the side, so that it’s swinging toward and away from you. Now let’s imagine that the string length, the masses of the weights, and the spring constant, etc., are all chosen so that the period of the oscillation is the same for both, and that the maximum and minimum heights of each weight is the same. From your perspective, they should look the same. Each weight should rise and fall in lock-step.

Now, when the spring-ball is at the bottom, it is easy to think that the tension on the spring must be greater than mg. If the tension were only mg, then the ball would sit motionless at the bottom. The added potential energy comes from the stretching of the spring, and this potential energy will convert to kinetic energy in lifting the ball up again. When it gets to the top, the potential energy comes from its height. This is how an oscillation works: conservation of energy, which converts from one form, to another, then back. In the case of the spring, the energy converts from potential (height) into kinetic (falling down) to potential (of the spring tension) to kinetic (the ball springing back up) and back to the start. This is a bit easier to understand intuitively than the pendulum, but it’s the same principle.

With the pendulum, let’s imagine that the center of rotation is the center of a clock, where the pendulum swings from eight o’clock, down to six, up to four, and then reverses back to six, and up to eight. At eight o’clock and four o’clock, the ball has no kinetic energy–it pauses motionless for an infinitessimally short moment. All of its energy is potential, from the height. At six o’clock, there is no potential energy. All of its energy is kinetic, as it reaches its maximum speed. At the time that the weight is at the bottom or the arc, the direction of the velocity is perpendicular to its arc. The string has the job of redirecting that kinetic energy to make the ball go up to either 8 or 4 o’clock. To do this, the direction of acceleration is perpendicular to the direction of the motion. This is centripetal force. Now, if you imagine a ball hanging motionless from a string, it has no kinetic energy. The total tension on the string in this situation is just -mg, as it is at equilibrium. The swinging pendulum at the bottom of its arc is not at equilibrium at all. Specifically, it has the same force of gravity as the ball at rest, but since the ball in the pendulum is swinging, it also has the force created by the acceleration–that is, the change in direction of the velocity–that comes at the bottom.

I think the key here is to remember that acceleration is not just the change in the speed, but the change in velocity. When you first hear about speed and velocity, it seems like a nit-pick to say that they are not the same, and that velocity is speed + direction. But a point on a wheel spinning at constant speed still accelerates. When you see that F = m*a, remember that acceleration doesn’t care if it’s a change in speed or a change in direction–it creates a force all the same. Since the pendulum ball changes direction of velocity, it must necessarily be accelerating, and since it has mass, there must be a resultant force. If the only tension on the string when the ball was at the bottom was -mg, then the system would be at equilibium, and the ball would be completely motionless.