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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F_c = \frac{m v^2}{r}\] | The rotating sphere of mass \(m\) needs a centripetal force \(F_c\) directed toward point \(P\) to keep it moving in a circle of radius \(r\). |
| 2 | \[F_s = k \Delta x\] | The only horizontal force capable of providing this inward pull is the spring force \(F_s\), where \(k\) is the spring constant and \(\Delta x\) is the extension beyond the natural length. |
| 3 | \[F_s = F_c\] | The spring stretches until its restoring force equals the required centripetal force, so the sphere moves outward and \(\Delta x\) increases while rotating. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[k = \frac{F}{\Delta x}\] | Hooke’s law gives the spring constant. The spring needs \(18.0\,\text{N}\) to stretch \(1.0\,\text{cm}=0.010\,\text{m}\). |
| 2 | \[k = \frac{18.0\,\text{N}}{0.010\,\text{m}} = 1.80 \times 10^3\, \text{N\,m}^{-1}\] | Calculate \(k\) numerically. |
| 3 | \[\Delta x = 0.265\,\text{m} – 0.250\,\text{m} = 0.015\,\text{m}\] | Find the new extension while the sphere rotates. |
| 4 | \[F_s = k \Delta x = (1.80 \times 10^3) (0.015) = 27\,\text{N}\] | Compute the spring force that acts as the centripetal force. |
| 5 | \[F_s = \frac{m v^2}{r}\] | Set the spring force equal to the centripetal force; here \(r = 0.265\,\text{m}\). |
| 6 | \[v^2 = \frac{F_s r}{m}\] | Algebraically solve for the square of the speed \(v\). |
| 7 | \[v^2 = \frac{27\, (0.265)}{0.075} = 95.4\,\text{m}^2\text{/s}^2\] | Insert numerical values using \(m = 0.075\,\text{kg}\). |
| 8 | \[v = \sqrt{95.4} = 9.77\,\text{m/s}\] | Take the square root to obtain the speed. |
| 9 | \[\boxed{v = 9.77\,\text{m/s}}\] | Final numerical answer, boxed as required. |
Just ask: "Help me solve this problem."
Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
A coffee cup on the dashboard of a car slides forward when the driver decelerates from \(45 ~ \frac{\text{km}}{\text{hr}}\) to rest in \(3.5 \, \text{s}\) or less. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard are completely horizontal.
A ball of mass \( m \) is fastened to a string. The ball swings at constant speed in a vertical circle of radius \( R \) with the other end of the string held fixed. Neglecting air resistance, what is the difference between the string’s tension at the bottom of the circle and at the top of the circle?
A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a wet patch in the road decreases the maximum static frictional force to one-third its dry road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
An object is moving to the west at a constant speed. Three forces are exerted on the object. One force is \( 10 \) \( \text{N} \) directed due north, and another is \( 10 \) \( \text{N} \) directed due west. What is the magnitude and direction of the third force if the object is to continue moving to the west at a constant speed?
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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] |
General Metric Conversion Chart
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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