Mike Quinn

Comment MCAT if you’re interested in the summer bootcamp. (feel free to dm me as well)

It will run from 5/26 - 7/23

MCAT physics - work done by gravity

Part II (go watch part I if you haven’t seen it)

The MCAT is going to ask you to calculate the work of gravity as an object is either lifted up, or brought towards the ground.

In this video we are calculating the work done by gravity to bring the ball back towards the ground.

Again, make sure the question is asking about the work done by GRAVITY. Other things can do work too, so make sure it’s gravity they’re asking about to use this method.

Work = Fdcos(theta)

“F” is for Force.

IMPORTANT - When you’re solving for the work done by gravity, you ALWAYS use mg to calculate F.

In this problem, mass is 5kg. g = 10m/s^2 on Earth.

“d” is for distance. In this video, the distance is 10m

“(theta)” is the angle created by the direction the object travels and the force acting on it.

This is a tricky concept. In this problem our object is traveling downwards. The force that we care about here is gravity, and it’s ALSO pointing downwards. The direction our object is traveling in and the force acting on it are THE SAME. This creates an angle of 0°.

cos (0°) = 1

Plugging in:

W = (5kg)(10m/s^2)(10m)cos(0°)= +500J

It’s important that you know why work is positive here. This indicates that gravity is responsible for bringing the ball towards the ground. It makes sense, gravity brings things downwards.

MCAT physics - work done by gravity

The MCAT is going to ask you to calculate the work of gravity as an object is lifted up.

Make sure the question is asking about the work done by GRAVITY. Other things can do work too, so make sure it’s gravity they’re asking about to use this method.

Work = Fdcos(theta)

“F” is for Force.

IMPORTANT - the MCAT can trick you by telling you there is a force applied to lift the object. This is NOT the force you plug into the formula. When you’re solving for the work done by gravity, you ALWAYS use mg to calculate F.

In this problem, mass is 5kg. g = 10m/s^2 on Earth.

“d” is for distance. In this video, the distance is 10m

“(theta)” is the angle created by the direction the object travels and the force acting on it.

This is tricky. In this problem our object is traveling upwards. The force that we care about here is gravity, and it’s pointing downwards. The direction our object is traveling in and the force acting on it are opposite. This creates an angle of 180°.

cos (180°) = -1

Plugging in:

W = (5kg)(10m/s^2)(10m)cos(180°) = -500J

It’s important that you know why work is negative here. This indicates that gravity is NOT lifting the ball up. It makes sense, gravity doesn’t lift things up.

This means that 500J of work was put in by something else to lift the ball. It could be a person, a machine, etc. that they give you in the problem.

MCAT Physics - Horizontal launch off a cliff

For a question like this you’re most likely to get a ball being launched off a cliff with a horizontal velocity.

It’ll be your job to calculate the distance the ball travels before reaching the ground.

Do it in 2 steps

1) find time

Use the equation shown in the video to calculate the time the ball spends in the air

2) find distance

Once you have time, multiply the horizontal velocity by time to get your answer.

MCAT Physics - Incline ramp

Make sure you know mg Sin(theta) & mg Cos (theta)

Practice problems to come

MCAT physics - the classic “incline ramp” question

In this video we cover the free body diagram.

Make sure you know how to label this free body diagram… also make sure you know how to derive the “fake” triangle that is drawn within the ramp. Part II to come.

MCAT Physics - Buoyant force

If you get a question asking about an object submerged in water, make sure you know the following:

The object will have two forces acting on it, both in the y direction. The downward force is the force due to gravity (mg). The upward force is called the “buoyant force.” This is the force of the water pushing upwards on the object.

Whenever you get a problem like this, you want to set up an equation for the sum of the forces in the y direction. Of note, your mg force will be negative because it points downwards and your buoyant force will be positive because it points upwards.

Our equation should be:

Sum of Forces (y-direction) = Fb - mg

In this video, our block is NOT accelerating, therefore there is NO net force acting on the block. Remember, Force = mass x acceleration. In this case, F = 0

Our force equation simplifies to:

0 = Fb - mg

Solving for our mass:

mg = Fb

m(10m/s^2) = 10N

mass = 1kg

MCAT physics - Hooke’s Law

You’re bound to get a spring question on the MCAT so make sure you know the following:

If no force is applied to the spring, it will stay at its equilibrium point. To move the spring from equilibrium, we can either compress it or expand it.

When you deviate the spring from its equilibrium point, it will exert a force to return to equilibrium. If you compress the spring to the left, it will try to go back to the right. If you expand the spring to the left, it will try to go back to the right.

In this problem the spring is displaced 5 meters to the left (compressed) this means our value for “x” is -5

Our spring has a spring constant of 100 N/m.

Solving for the force of our spring:

F = - 100N/m x (-5m) = 500N

MCAT Orgo

For any question regarding a polarimeter, make sure know the following:

If you put a CHIRAL molecule in a polarimeter, it will “rotate plane-polarized light.” Again, only chiral molecules do this. You don’t need to know how a polarimeter works. Just know that only chiral molecules can rotate the light.

This rotation of light gives the molecule an “optical rotation”. The value can be positive or negative.

Big idea:
Two molecules that have equal and opposite optical rotations (e.g +25 & -25) are enantiomers

Two molecules that have unrelated rotations (e.g +40 & -25) are diastereomers

If the question mentions a 50:50 racemic mixture of enantiomers, just know the optical rotation is 0

MCAT physics - Pascal’s Law

Pascal’s law says that the pressure applied to a fluid is uniform throughout. In other words, it doesn’t matter where you apply pressure to the fluid… the entire fluid feels the exact same pressure

For this reason, we can say that Pascal‘s law is as follows:

P1 = P2

This simply means that the pressure at one point in the fluid is equal to the pressure at another point in the fluid.

Pressure = (Force/Area)

To solve our problem in the video, we plug in (F1/A1) = (F2/A2)