| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$F_{\parallel} = F\cos(\theta)$$ | The horizontal force \( F \) is resolved into a component along the inclined plane. The component along the plane is given by \( F\cos(\theta) \). |
| 2 | $$F_{g\,\parallel} = mg\sin(\theta)$$ | The gravitational force has a component down the incline given by \( mg\sin(\theta) \). |
| 3 | $$N = mg\cos(\theta)+ F\sin(\theta)$$ | The normal force \( N \) is found by resolving forces perpendicular to the incline. Gravity gives \( mg\cos(\theta) \) and the horizontal force contributes \( F\sin(\theta) \) pushing the block into the plane. |
| 4 | $$F_{f} = \mu N = \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)$$ | The frictional force is given by the coefficient of friction \( \mu \) times the normal force. |
| 5 | $$m\,a = F\cos(\theta)- mg\sin(\theta)- \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)$$ | Applying Newton’s second law along the incline, the net force is the sum of the component of \( F \) along the plane minus both the gravitational and frictional forces. |
| 6 | $$a = \frac{F\cos(\theta)- mg\sin(\theta)- \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)}{m}$$ | This is the final expression for the block’s acceleration \( a \) up the incline in terms of \( m,\,\theta,\,\mu,\,F, \) and \( g \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$0 = F\cos(\theta)- mg\sin(\theta)- \mu\left(mg\cos(\theta)+ F\sin(\theta)\right)$$ | For the block to slide up the plane with constant velocity, the acceleration must be zero. Hence the net force along the incline is zero. |
| 2 | $$F\cos(\theta)- \mu F\sin(\theta) = mg\sin(\theta)+ \mu mg\cos(\theta)$$ | Rearrange the equation by grouping the terms involving \( F \) on the left and the gravitational terms on the right. |
| 3 | $$F \left(\cos(\theta)- \mu \sin(\theta)\right) = mg\left(\sin(\theta)+ \mu\cos(\theta)\right)$$ | Factor out \( F \) on the left-hand side and \( mg \) on the right-hand side for clarity. |
| 4 | $$F = \frac{mg\left(\sin(\theta)+ \mu\cos(\theta)\right)}{\cos(\theta)- \mu\sin(\theta)}$$ | Solve the equation for \( F \) by dividing both sides by \(\cos(\theta)- \mu \sin(\theta)\). |
| 5 | $$\cos(\theta)- \mu\sin(\theta)> 0 \quad \Longrightarrow \quad \tan(\theta) < \frac{1}{\mu}$$ | For \( F \) to be physically meaningful (i.e., a positive real number), the denominator must be positive. Rearranging the inequality yields the condition \( \tan(\theta) < \frac{1}{\mu} \). |
| 6 | $$\boxed{F = \frac{mg\left(\sin(\theta)+ \mu\cos(\theta)\right)}{\cos(\theta)- \mu\sin(\theta)}}$$ | This is the final expression for the magnitude of the applied horizontal force required to make the block slide with a constant velocity, including the physical condition on \( \theta \) and \( \mu \). |
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A loop-de-loop roller coaster has a radius of \( 30 \) \( \text{m} \). Determine the apparent weight a \( 500 \) \( \text{N} \) person will feel at the bottom of the loop while traveling at a speed of \( 25 \) \( \text{m/s} \) and at the top of the loop while traveling at a speed of \( 20 \) \( \text{m/s} \).
Two objects, \( A \) and \( B \), move toward one another. Object \( A \) has twice the mass and half the speed of object \( B \). Which of the following describes the forces the objects exert on each other when they collide and provides the best explanation?
Suppose you place a ball in the middle of a wagon, and then accelerate the wagon forward. Describe the motion of the ball relative to the ground. Describe its motion relative to the wagon.
If I weigh \( 741 \) \( \text{N} \) on Earth at a place where \( g = 9.80 \) \( \text{m/s}^2 \) and \( 5320 \) \( \text{N} \) on the surface of another planet, what is the acceleration due to gravity on that planet?
A spring is connected to a wall and a horizontal force of \( 80.0 \) \( \text{N} \) is applied. It stretches \( 25 \) \( \text{cm} \); what is its spring constant?
When the speed of a rear-wheel-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
The figure shows a truck pulling three crates across a rough road. Which of the following best describes the directions of all the horizontal forces acting on crate 2?
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.
A person pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?
What force would have to be applied to start a \(12.3 \, \text{kg}\) wood block moving on a surface with a static coefficient of friction of \(0.438\)?
a) \( a = \frac{F (\cos \theta – \mu \sin \theta) – mg (\mu \cos \theta + \sin \theta)}{m} \)
b) \( F = \frac{\mu m g \cos \theta + \mu F \sin \theta + m g \sin \theta}{\cos \theta} \)
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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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