| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \frac{1}{2}m v_{0}^{2} = m g \Delta x \sin(34^\circ) \] | This is the energy conservation for a frictionless slide where all the gravitational potential energy \(m g \Delta x \sin(34^\circ)\) is converted into kinetic energy \(\frac{1}{2}m v_{0}^{2}\) at the bottom. |
| 2 | \[ v_{0} = \sqrt{2g \Delta x \sin(34^\circ)} \] | Solve for the frictionless final speed \(v_{0}\) by isolating it in the energy equation. |
| 3 | \[ \frac{1}{2}m v_{x}^{2} = m g \Delta x \sin(34^\circ) – \mu \; m g \Delta x \cos(34^\circ) \] | For a slide with kinetic friction, the work done by friction \(\mu m g \Delta x \cos(34^\circ)\) is subtracted from the available gravitational potential energy. |
| 4 | \[ v_{x} = \frac{1}{2}v_{0} \] | It is given that the child’s speed at the bottom with friction is exactly half the frictionless speed. |
| 5 | \[ \frac{1}{2}m \left(\frac{1}{2}v_{0}\right)^2 = m g \Delta x \sin(34^\circ) – \mu \; m g \Delta x \cos(34^\circ) \] | Substitute \(v_{x} = \frac{1}{2}v_{0}\) into the energy equation with friction. |
| 6 | \[ \frac{1}{2} \left(\frac{1}{2}v_{0}\right)^2 = g \Delta x \sin(34^\circ) – \mu \; g \Delta x \cos(34^\circ) \] | Cancel the mass \(m\) from both sides since it appears throughout. |
| 7 | \[ \frac{1}{2} \left(\frac{1}{4}v_{0}^2\right) = \frac{1}{8}v_{0}^2 = g \Delta x \sin(34^\circ) – \mu \; g \Delta x \cos(34^\circ) \] | Simplify the left side by computing \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and then multiplying by \(\frac{1}{2}\). |
| 8 | \[ \text{Since} \; \frac{1}{2}v_{0}^2 = g \Delta x \sin(34^\circ), \; \text{we have} \; \frac{1}{8}v_{0}^2 = \frac{1}{4}g \Delta x \sin(34^\circ) \] | Replace \(\frac{1}{8}v_{0}^2\) using the frictionless energy equation for consistency. |
| 9 | \[ \frac{1}{4}g \Delta x \sin(34^\circ) = g \Delta x \sin(34^\circ) – \mu \; g \Delta x \cos(34^\circ) \] | Equate the expression obtained from energy with friction to the simplified form of frictionless energy. |
| 10 | \[ \frac{1}{4}\sin(34^\circ) = \sin(34^\circ) – \mu \cos(34^\circ) \] | Cancel \(g \Delta x\) from both sides since they are nonzero. |
| 11 | \[ \sin(34^\circ) – \frac{1}{4}\sin(34^\circ) = \mu \cos(34^\circ) \] | Simplify the right side by subtracting \(\frac{1}{4}\sin(34^\circ)\) from \(\sin(34^\circ)\). |
| 12 | \[ \frac{3}{4}\sin(34^\circ) = \mu \cos(34^\circ) \] | This gives the relationship that relates \(\mu\) to the sine and cosine of \(34^\circ\). |
| 13 | \[ \mu = \frac{\frac{3}{4}\sin(34^\circ)}{\cos(34^\circ)} = \frac{3}{4}\tan(34^\circ) \] | Solve for \(\mu\) by dividing both sides by \(\cos(34^\circ)\). |
| 14 | \[ \boxed{\mu \approx 0.51} \] | Substitute \(\tan(34^\circ) \approx 0.67\) to get a numerical value \(\mu \approx \frac{3}{4} \times 0.67 \approx 0.50-0.51\). This is the coefficient of kinetic friction. |
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A \(30 \, \text{g}\) bullet is fired with a speed of \(500 \, \text{m/s}\) into a wall.
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
The coefficient of static friction between hard rubber and normal street pavement is about \(0.85\). On how steep a hill (maximum angle) can you leave a car parked?
A comet of mass \( m_c = 3.2 \times 10^{14} \) \( \text{kg} \) is orbiting a star with mass \( m_s = 1.8 \times 10^{30} \) \( \text{kg} \). The comet’s orbit is elliptical. At its closest point, the comet is a distance \( r_1 = 8.3 \times 10^{10} \) \( \text{m} \) from the star, and at its farthest point, the comet is a distance \( r_2 = 4.9 \times 10^{11} \) \( \text{m} \) from the star. What is the change in the kinetic energy of the comet as it moves along its orbit from distance \( r_2 \) to distance \( r_1 \) from the star?
Find the net gravitational force on a \(2.0 \, \text{kg}\) sphere midway between a \(4.0 \, \text{kg}\) sphere and a \(7.0 \, \text{kg}\) sphere that are \(1.2 \, \text{m}\) apart.
When a falling meteoroid is at a distance above the Earth’s surface of \( 3.00 \) times the Earth’s radius, what is its acceleration due to the Earth’s gravitation?
An elastic cord is \( 80\) \( \text{cm} \) long when it is supporting a mass of \( 10. \) \( \text{kg} \) hanging from it at rest. When an additional \( 4.0 \) \( \text{kg} \) is added, the cord is \( 82.5 \) \( \text{cm} \) long.
On a harsh winter day, a \( 1500 \) \( \text{kg} \) vehicle takes a circular banked exit ramp (radius \( R = 150 \) \( \text{m} \); banking angle of \( 10^\circ \)) at a speed of \( 30 \) \( \text{mph} \), since the speed limit is \( 35 \) \( \text{mph} \). However, the exit ramp is completely iced up (frictionless). To make matters worse, a wind is blowing parallel to the ramp in a downward direction. The wind exerts a force of \( 3000 \) \( \text{N} \). Under these conditions, can the driver continue to follow a safe horizontal circle on the exit ramp and stay below the speed limit?
To convert \( \text{mph} \) into \( \text{m/s} \), use \( 1 \) \( \text{mi} = 1607 \) \( \text{m} \) and \( 1 \) \( \text{hr} = 3600 \) \( \text{s} \).
In an experiment where a constant horizontal force pulls on a box across a rough floor starting from rest, what would happen to the acceleration of the box if its mass were doubled but the pulling force remained unchanged?
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.
\(\boxed{\mu \approx 0.51}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\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}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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