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MCAT Physics > 2: Newtonian Mechanics > Flashcards

2: Newtonian Mechanics Flashcards

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1
Q

Force

A

(F) is a vector quantity that is experienced as pushing or pulling on objects

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2
Q

SI unit of Force

A

Newton (N) = kg*m/s^2

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3
Q

Mass

A

(m) Measure of a body’s inertia - the amount of matter in something
- Scalar

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4
Q

SI unit of mass

A

kg

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5
Q

Weight

A

(W) measure of gravitational force, usually that of earth, on an object’s mass
-Vector

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6
Q

SI unit of weight

A

Newton (N)

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7
Q

W=

A

mg

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8
Q

Acceleration (a)

A

The rate of change of velocity that an object experiences as a result of some applied force
-Vector

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9
Q

SI unit of acceleration

A

m/s^2

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10
Q

Deceleration

A

acceleration in the direction opposite the initial velocity

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11
Q

Body in motion, stays in motion

A

F=ma = 0

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12
Q

2nd law

A

Sum F = ma

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13
Q

Equal and oppositie

A

Fb = -Fa

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14
Q

Gravity ______ with height above the earth and __________ the closer you get to the earth’s center of mass.

A

Decreases

Increases

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15
Q

If there is no acceleration

A

Then there is no net force on the object

- This means that any object with a constant velocity has no net force acting on it

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16
Q

Net force

A

Sum of all forces acting on an object
- Even though the force of gravity is always acting on us, the net force on our bodies will be zero, unless there is no ground below us pushing back up against gravity

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17
Q

Gravity

A

An attractive force that is felt by all forms of matter

- Weakest of the four types of forces

18
Q

The magnitude of the gravitational force is

A

Inverse to the square of the distance

- That is, if r is halved, the F will quadruple

19
Q

F =

A

Gm1m2/ r^2
- Where G is the universal gravitational constant (6.67x10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two objects and r is the distance between their centers

20
Q

Translational motion

A

When forces cause an object to move without any rotation about a fixed point in the object
- Common on the MCAT

21
Q

Rotational motion

A

When forces are applied against an object in such a way as to cause the object to rotate around a fixed pivot point, also known as the fulcrum
- Generate torque and moment of force

22
Q

Torque

A

Generates the rotational motion, not the mere application of force itself
- This is because torque depends not only on the magnitude of the force but also on the angle at which the force is applied against the lever arm as well as the distance between the fulcrum and the point of force application
- T = rF sin theta
Where F is the magnitude of force, r is the distance between the fulcrum and the point of force application, and theta is the angle between F and the lever arm

23
Q

Circular motion

A

When forces cause an object to move in a circular pathway

  • Upon completion of one cycle, the displacement of the object is zero
  • F = mv^2/r
24
Q

For circular motion that demonstrates a constant speed at all points along the pathway

A

The instantaneous velocity vector is always tangent to the circular path

25
Q

Uniform circular motion

A

The tangential force is zero (because there is no change in the speed of the object). Thus the resultant force is the radial force. This is known as the centripetal force, which generates centripetal acceleration

26
Q

Centripetal acceleration

A

Keeps an object in its circular pathway

  • When the centripetal force is no longer acting on the object it will simply exit the circular pathway and assume a path tangential to the circle at that point
  • v^2/r
27
Q

Nonuniform circular motion

A

The speed of the moving object changes over the course of the path. This means that there is a tangential force acting to create a tangential acceleration. This force vector adds to the radial force vector to produce a resultant force (and resultant acceleration) that is not directed toward the center of the circle.
- On MCAT, these will have to do with planetary and satellite orbits

28
Q

Friction

A

Kind of force that works to oppose the movement of objects

29
Q

Static Friction (Fs)

A

Exists between a stationary object and the surface upon which it rests
0< (or equal to) fs< (or equal to) Mu s Fn
- Mu s is the coefficient of static friction
- Fn is the normal force (don’t forget that the normal force is the component of the contact force that is perpendicular to the plane of contact between the object and the surface upon which it rests)
- A range of possible values for static friction

30
Q

Contact points

A

The places where friction occurs between two rough surfaces sliding past each other

31
Q

Kinetic Friction (Fk)

A

Exists between a sliding object and the surface over which the object slides

  • Any time two objects slide against each other, kinetic friction will be present
  • fk = Mu k Fn
  • Kinetic friction will have a constant value for any given combination of coefficient of kinetic friction and normal force
32
Q

The maximum value for static friction will always be

A

Greater than the constant value for kinetic friction

- Fs > Fk

33
Q

Pay close attention to

A

The conditions set in the MCAT passage or question. Does it tell you that friction can be assumed negligible, or does it provide the coefficient values, which will most likely need to be used in a calculation of friction values?

34
Q

Mechanical equilibrium

A

Occurs when the vector sums of the forces or torques acting on an object is zero (all of the force or all of the torque vectors cancel out)

35
Q

Translational equilibrium

A

First Condition
Exists only when the vector sums of all of the forces acting on an object is zero
- When the resultant force upon an object is zero, the object will not accelerate. Its motional behavior will be constant.
- What is important to remember is that an object experiencing translational equilibrium will have a constant speed (which could be zero) and a constant direction

36
Q

Rotational equilibrium

A

Second Condition

Exists only when the vector sum of all the torques acting on an object is zero (all positives cancel out all negatives)

37
Q

Negative Torques

A

Clockwise rotation

38
Q

Positive Torques

A

Counterclockwise rotation

39
Q

Two possibilities of motion in the case of rotational equilibrium

A
  1. The lever arm is not rotating at all (it’s stationary)
  2. The lever arm is rotating with a constant angular frequency (analogous to constant velocity)
    - MCAT usually uses the stationery lever arm
40
Q

REMEMBER

A

Since sin of 90 degrees equals 1, this means that torque is greatest when the force applied is 90 degrees (perpendicular) to the length of the lever arm. Knowing that sin of 0 degrees equals 0 tells us that there is no torque when the force applied is parallel to the lever arm

MCAT Physics (2 decks)

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