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Step Derivation/Formula Reasoning
1 KE_{\text{peak}} = \frac{1}{2} m v^2 Calculate the kinetic energy at the peak height. Kinetic energy (KE) is given by the equation \frac{1}{2} m v^2, where m is mass (2 kg) and v is velocity (150 m/s).
2 PE_{\text{peak}} = mgh Calculate the potential energy at the peak height. Potential energy (PE) is given by mgh where m is mass (2 kg), g is the acceleration due to gravity (approximately 9.8 \, \text{m/s}^2) and h is the height (90 m).
3 W_{\text{thrust}} = F_{\text{thrust}} \times h Calculate the work done by the thrust force. Work (W) is force (F) times the displacement (h). Here, F_{\text{thrust}} is 275 N.
4 KE_{\text{peak}} = \frac{1}{2} \times 2 \times 150^2 = 22500 \, \text{J} Substitute the values to find KE_{\text{peak}} .
5 PE_{\text{peak}} = 2 \times 9.8 \times 90 = 1764 \, \text{J} Substitute the values to find PE_{\text{peak}} .
6 W_{\text{thrust}} = 275 \times 90 = 24750 \, \text{J} Substitute the values to find W_{\text{thrust}} .
7 W_{\text{thrust}} = KE_{\text{peak}} + PE_{\text{peak}} + W_{AR} Use the conservation of energy. In this case the initial work done by the rocket thrust transforms into kinetic energy, potential energy, and work done by air resistance W_{AR} .
8 W_{AR} = 24750 – 22500 – 1764 Solving for the work done by air resistance gives us W_{AR} = 486 \, \text{J} .
9 W_{AR} = Fh Divide the work done by air resistance by the height to find the average air resistance force.
10 W_{AR} = \frac{486}{90} \approx 5.4 \, \text{N} F_{\text{air}} \approx \boxed{5.4} \, \text{N} is the average air resistance force acting against the rocket during ascent.

ALTERNATE EXPLANATION – Using Forces

Step Derivation/Formula Reasoning
1 F_{\text{net}} = ma Newton’s second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
2 F_{\text{net}} = F_{\text{thrust}} – F_{\text{gravity}} – F_{\text{drag}} The net force acting on the rocket includes the thrust force, the gravitational force, and the drag force (air resistance), where the thrust and drag act upwards, and gravity acts downwards.
3 F_{\text{gravity}} = mg The gravitational force is calculated as the product of the mass and the acceleration due to gravity g \approx 9.81 \, \text{m/s}^2 .
4 F_{\text{gravity}} = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} Substitute the values of the mass and gravitational acceleration to find the gravitational force.
5 a = \frac{v^2 – v_0^2}{2h} The kinematic equation relates the rocket’s change in velocity to acceleration and distance. Here, v_0 = 0 \, \text{m/s} (initial velocity), v = 150 \, \text{m/s} (final velocity at the peak), and h = 90 \, \text{m} (height).
6 a = \frac{(150 \, \text{m/s})^2}{2 \times 90 \, \text{m}} Substituting the values into the kinematic equation to calculate acceleration.
7 a = 125 \, \text{m/s}^2 Calculation of the acceleration using the given values.
8 F_{\text{net}} = 2 \, \text{kg} \times 125 \, \text{m/s}^2 = 250 \, \text{N} Calculate the net force acting on the rocket using Newton’s second law with the rocket’s mass and the acceleration.
9 250 \, \text{N} = 275 \, \text{N} – 19.62 \, \text{N} – F_{\text{drag}} Solve for the drag force using the equation in step 2.
10 F_{\text{drag}} = 275 \, \text{N} – 19.62 \, \text{N} – 250 \, \text{N} = 5.38 \, \text{N} Substitute the values of the thrust, gravitational, and net forces to find the drag force.
11 F_{\text{drag}} \approx 5.38 \, \text{N} This is the average air resistance force acting on the rocket during its ascent.

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F_{\text{air}} \approx 5.4 \, \text{N}

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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