Video Transcript: Introduction to Forces in MCAT Physics

If you prefer to watch it, see the video HERE, or catch the entire Forces in MCAT Physics series HERE

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[Start Transcript]

Leah here from leah4sci.com/MCAT and in this video series I’ll give you very detailed foundation on basics that you need to know on Forces on the MCAT. You can find my entire video of series along with my forces cheat sheets and MCAT practice quiz by visiting my website leah4sci.com/Forces.

When you think of Forces, the standard equation that comes to mind is F=ma. When we think of Forces, I picture an object that is being pushed or pulled and starts to move or slows down and that directly applies to the equation because the object in question is the mass and causing it speed up or slow down comes from change of acceleration positive or negative and that’s how we get F = ma.

But when the average student studies for the MCAT they memorize instead of understanding equations and then they have trouble applying it to the passages in the MCAT style questions. Instead what you need to do is take every equation not just for forces but every equation and break it down into the units and practice deriving every unit. So let’s apply this to forces.

A force is measured in Newton which is equals to kilogram times meter per second squared and we see this in equation F equals ma because mass is measured in kilograms and acceleration is measured in meters per second squared. But don’t stop here, if we have acceleration that means we can type forces into the kinematic equations and say that acceleration is equal to velocity over time so Force is equals to mass times velocity over time. Velocity is measured in meters per second divided by time it is per second, which is we have to have seconds on the denominator and that ones again gives me meters per second squared.  So if Force is equals to mass times velocity over time we ones again get kilograms meter per second squared but you shouldn’t stop here. If F is equals to ma, practice solving for a and prove that the units work out. To isolate a we divide both sides by m which gives us a new equation which is a is equals to F over m. If is measured in Newton which is kilogram times meter per second squared and mass is measured in kilograms, the kilograms canceled leaving us with meter per second squared which is the correct unit for acceleration. This is how you actively practice every equation to make sure you get it, you know how to derive it and in doing so you’ll memorize it based on logic rather than just a whole lot of random equations that don’t make any sense.

If you’re not comfortable with Kinematic equations visit my website leah4sci.com/MCATPhysics and find my entire series and cheat sheets on kinematics to make sure that you get the foundation before moving on to the rest of the forces series.

Let’s try a simple problem. Here we’re told to find the force on a 1.23 times ten to the third kilogram car when it accelerates at 3.9 meters per second squared. We’re told to find the Force so we know we’re solving for Newton and we can use the standard equation F equals ma. In this case the mass is 1.23 times ten to the third kilograms and the acceleration is 3.9 meters per second squared. First thing you want to do is to look the units make sure it works out. Force or Newtons is measured in kilogram times meters per second squared so we know it’s setup correctly but now we have to do the math without a calculator. 1.23 times ten to the third times 3.9 is kinda daunting so you wanna simplify the numbers and round it because remember on the MCAT, close enough is good enough. So we round the mass  1.2 times ten to the third and the acceleration to four. So now we have 1.2 times four. One times four is four, two times four is eight, gives me 4.8 keep the exponent times ten to the third and the unit’s Newton. So the force is 4.8 times 10 to the third newtons. To show you how close we are, on the calculator I got 4.797 times ten to the third which is actually 4.8 when you round it.

Now let’s take this a step further and see how to connect Forces with Kinematics in this problem.

A force of negative twelve Newtons is applied to a two thousand three hundred fifty grams toy car moving at eight meters per second. How far does it travel before coming to a complete stop?

Let’s take a look at what we have. A force of negative twelve Newtons implies that the force is going opposite the direction of motion so we’re slowing it down and that’s why it comes to a complete stop. We have an initial velocity of eight meters per second and if it comes to a complete stop we have a final velocity of zero meters per second. We’re asked to find how far it travels which means delta x are unknown. We almost have enough information to apply  the kinematic equations except that the twelve newtons doesn’t plugin we have to exact the acceleration.

So with F is equal to ma which is equal to negative twelve newtons we are going to divide both sides by the mass and that will give us acceleration. So acceleration is equal to negative twelve newtons divided by the mass which has to be in kilograms and not grams. 2350 grams has to be divided by a thousand to get kilograms or simply move the decimal back three spaces or 2.35 which we can round to 2.4 kilograms. The reason we round to 2.4 is 12 nearly divides nicely by 2.4. We’ll use the decimal trick here to move the top and bottom one over to the right. If you’re not comfortable with this, go back to my MCAT Math series on my website leah4sci.com/MCATMath. This gives us a hundred and twenty which we’ll just look at as twelve times ten over twenty four.  Twelve is half of twenty four giving me one over two and ten over two is equal to five, don’t forget the negative sign  which gives us an acceleration of negative five meters per second squared. Negative implies it’s opposite the direction of motion.

We’ll add negative five meters per second squared to our list of equations and now go back to the Kinematic concept of which equation do we use?  When we have initial final velocity, delta x, and acceleration, we want to use the time independent equation that says V final squared is equals to V initial squared plus 2a delta x. since the final velocity is zero, we’ll cross that out and then start to isolate the delta x which is what we are trying to solve. We’ll bring the initial velocity over to the other side giving us the equation negative initial velocity squared is equal to two a delta x and then we’ll divide by 2a isolating delta x.  Delta x is equal to negative V initial squared over 2a so let’s go ahead and plug it in. we have negative times eight meters per second squared divided by two times negative five meters per second squared. The negatives cancel out so that we’re going to have a positive value and then we’ll just have to do the math. Eight squared to sixty four divided by two times five which is ten. Sixty four over ten means we can use the factor of ten trick just move the decimal back one space for an answer of six point four, now let’s take a look at the units.

Meters per second squared means that we have meters over meters per second so let’s distribute that into the equation. The seconds square cancels and the meters square we only cancel one because we only have one meter in the denominator which leaves me with the final unit of meters which is correct for a delta x calculation.

In the upcoming videos I’ll cover Newton’s laws and then the various forces that you have to know including Normal Force, Gravitational, Tension, Static and Kinetic, Friction, Centrifugal Force, and Rotational such as Torque.

You can find this entire series along with my Forces cheat sheet along with my practice quiz by visiting my website at leah4sci.com/Forces.

Are you stuck on a specific MCAT topic? I offer Private Online Tutoring where I focus on your needs to strengthen your individual weaknesses. Tutoring details can be found using the link below or by visiting my website leah4sci.com/MCATTutor.

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Click here to Catch my entire video series on MCAT Forces