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}}. |
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 |
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. |
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}}. |
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:
Phy can also check your working. Just snap a picture!
While flying horizontally in an airplane, you notice that a string dangling from the overhead luggage compartment hangs at rest at 15° away from the vertical toward the back of the plane. Using this observation, you can conclude that the airplane is:
Shown above are three masses of 6 kg, 3 kg, and 1 kg (from left to right). You pull on the first mass with a force of 15 N along a frictionless surface.
Why do raindrops fall with constant speed during the later stages of their descent?
A truck of mass 3500 kg hits the back of a small car of mass 1400 kg. Which car exerted more force on the other and why?
A small sphere hangs from a string attached to the ceiling of a uniformly accelerating train car. It is observed that the string makes an angle of 37° with respect to the vertical. The magnitude of the acceleration a of the train car is most nearly:
6.67 seconds
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Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_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 Motion | Energy |
---|---|
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 |
Momentum | Torque and Rotations |
---|---|
p = m v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum mr^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients 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. |
k | Spring 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 |
Variable | SI 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)} |
Variable | Derived 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)} |
\omega (Angular Velocity) | \text{radians per second (rad/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.
Start with the given measurement: \text{5 km}
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}}
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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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