Motion and Forces II - Detailed Notes

2.1 Acceleration

• Acceleration $(a)$ is a vector describing the rate of change of velocity over time, considering both magnitude and direction.

• Units for acceleration are meters per second squared $(m/s^2)$.

• Formula: $a = \frac{\Delta v}{t}$, where $\Delta v$ represents the change in velocity and $t$ is the time interval.

• $\Delta v = v<em>f - v</em>i$, where $v<em>f$ is the final velocity and $v</em>i$ is the initial velocity.

• On a velocity-time graph, the slope indicates the acceleration of the object.

• Average velocity for constant acceleration: $v<em>{avg} = \frac{1}{2} (v</em>i + v_f)$. This is the vertical midpoint of the line, valid only for constant acceleration.

• A curved line on a position-time graph signifies acceleration because the slope (velocity) is changing.

2.2 Solving Acceleration Problems

• Equations for constant acceleration:

  • $\Delta x = v_i t + \frac{1}{2} a t^2$

  • $v<em>f = v</em>i + a t$

  • $v<em>f^2 = v</em>i^2 + 2 a \Delta x$

• Problem-solving method (SOLVE):

  • Sketch: Draw a diagram and choose a positive direction.

  • Objective: Identify the variable to be solved for.

  • List Variables: List known and unknown variables.

  • Equation: Select the appropriate equation that relates the variables and solve for the unknown.

*Example Problem: A car traveling 20 m/s (W) comes to a stop with an acceleration of 0.8 m/s² (E). Calculate the distance traveled while stopping. $\Delta x = ?$; $v<em>i = 20 m/s$; $v</em>f = 0 m/s$; $a = -0.8 m/s^2$; Using $v<em>f^2 = v</em>i^2 + 2a\Delta x$; $0^2 = 20^2 + 2(-0.8) \Delta x$ resulting in $\Delta x = 250m$

2.3 Newton’s Second Law

• Newton’s Second Law relates force, mass, and acceleration.

• A smaller mass experiences a larger acceleration when acted upon by a force compared to a larger mass with the same force.

• The acceleration of an object is directly proportional to the net external force and inversely proportional to its mass.

• Mathematically: $a = \frac{F<em>{net}}{m}$ or $F</em>{net} = ma$

• $\sum F = ma$ (Net force equals mass times acceleration).

• The direction of acceleration is the same as the direction of the net force.

2.4 Friction

• Friction is a contact force that opposes the relative motion of two surfaces.

• Caused by microscopic interactions between surfaces.

• Coefficient of friction $(\mu)$ depends on the types of surfaces in contact.

• Two types of friction: static and kinetic.

• Static friction ($\mu<em>s$) applies to objects at rest, while kinetic friction ($\mu</em>k$) applies to moving objects.

• Static friction is generally greater than kinetic friction.

• Friction also depends on the normal force.

• Static Friction: $F<em>f \le \mu</em>s F<em>N$ (Static friction is less than or equal to the maximum value of $\mu</em>s F_N$).

• Kinetic Friction: $F<em>f = \mu</em>k F_N$

• Temperature increases due to friction.

• Air resistance is similar to friction.

2.5 Springs

• Spring Constant $(k)$: Describes the stiffness of a spring. It is the force required to stretch (or compress) a spring divided by the distance. Unit: N/m. A large $k$ means a big force is needed to stretch it.

• Formula: $F<em>s = kx$, where $F</em>s$ is the force required to compress the spring a distance $x$ meters from its equilibrium (rest) point.

2.6 Solving N2 Problems

*To use Newton’s 2nd Law of Motion
1. Draw a free-body diagram, labeling all forces (magnitudes and directions) with appropriate positive and negative signs.
2. Add all the forces in each direction. If the sum is zero, the forces are balanced, and there is no acceleration. If the sum is not zero, find the net force in each direction. Set the net force in each direction equal to $ma$ and solve for the unknown quantity.

*Example: $F_{NET} = ma$; $10000N - 8000N = 1200kg * a$; $a = 1.67 m/s^2$