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First find the pushing force, which equal to the maximum force of static friction.

Step Formula Derivation Reasoning
1 F_{\text{static max}} = \mu_{s} F_{\text{normal}} Maximum static friction force, where \mu_{s} is the coefficient of static friction.
2 F_{\text{normal}} = mg Normal force equals weight for horizontal motion.
3 F_{\text{static max}} = \mu_{s} mg Substituting F_{\text{normal}}.

#### Now find the net force during the first 5 seconds. Use this to find the acceleration of the block in the first 5 seconds.

Step Formula Derivation Reasoning
1 F_{\text{net}} = F_{\text{push}} – F_{\text{kinetic}} Net force is the difference between pushing force and kinetic friction.
2 F_{\text{push}} = \mu_{s} F_{\text{normal}} Pushing force equal to maximum static friction.
3 F_{\text{kinetic}} = \mu_{k} F_{\text{normal}} Kinetic friction force.
4 F_{\text{net}} = \mu_{s} mg – \mu_{k} mg Substituting values for F_{\text{push}} and F_{\text{kinetic}}.
5 F_{\text{net}} = \mu_{s} mg – \mu_{k} mg = ma Set net force equation equal to ma. Then solve for a (acceleration)
6 a = 1 m/s2 Plug in values and solve

#### Final Speed at End of 5 Seconds

Step Formula Derivation Reasoning
1 v = u + \frac{F_{\text{net}}}{m}t Kinematic equation for velocity. Note F/m is the acceleration (1 m/s2 ) from above.
2 v = 0 + \frac{(\mu_{s} – \mu_{k}) mg}{m} \times 5, \text{s} Initial velocity u = 0; substituting F_{\text{net}} and t.
3 v = 5 m/s Plug in values and find the velocity at the end of the 5 second push.

#### Time to Stop Due to Just Kinetic Friction (after you stop pushing)

Step Formula Derivation Reasoning
1 0 = v – \mu_{k} gt_{\text{stop}} Final velocity 0 when stopped, v is final speed after push (5 m/s as found above).
2 t_{\text{stop}} = \frac{v}{\mu_{k} g} Solving for time to stop t_{\text{stop}}.

### Total Time

Step Formula Derivation Reasoning
1 t_{\text{total}} = t_{\text{push}} + t_{\text{stop}} Sum of time with force applied and time to stop.

Let’s calculate t_{\text{total}}. The calculations yield the following results:

1. Final speed at the end of 5 seconds: \boxed{4.905, \text{m/s}}
2. Time to stop due to kinetic friction: \boxed{1.667, \text{seconds}}
3. Total time the block is in motion: \boxed{6.667, \text{seconds}}

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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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