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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Setting \( y = 0 \) at the compressed position of the spring | By setting \( y = 0 \) at the position where the spring is compressed, we simplify the calculations for the potential energy of the projectile when it is launched. This allows us to treat the initial potential energy as entirely elastic and the final height as purely gravitational potential energy. |
Next, let’s calculate the spring constant:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(EPE = \frac{1}{2}kx^2\) | Elastic Potential Energy (EPE) stored in the compressed spring. \( x \) is the compression distance. |
| 2 | \(GPE = mgh\) | Gravitational Potential Energy (GPE) at the maximum height \( h \) above the initial position. \( m \) is the mass, \( g \) is gravitational acceleration, and \( h \) is the height. |
| 3 | \(\frac{1}{2}kx^2 = mgh\) | Applying conservation of energy, the EPE at the beginning is converted to GPE at the maximum height. |
| 4 | \(k = \frac{2mgh}{x^2}\) | Solving for the spring constant \( k \). |
| 5 | \(k = \frac{2 (0.035 \, \text{kg}) (9.8 \, \text{m/s}^2) (25 \, \text{m})}{(0.120 \, \text{m})^2}\) | Substitute the known values: \( m = 0.035 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 25 \, \text{m} \), \( x = 0.120 \, \text{m} \). |
| 6 | \(k = \frac{2 \cdot 0.035 \cdot 9.8 \cdot 25}{0.0144}\) | Calculate the values inside the equation. |
| 7 | \(k = \frac{17.15}{0.0144} \approx 1191.67 \, \text{N/m}\) | Divide to find \( k \). Therefore, the spring constant is approximately \( \boxed{1191.67 \, \text{N/m}} \). |
Finally, let’s calculate the speed:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(EPE = \frac{1}{2}kx^2\) | Using the elastic potential energy stored in the spring. |
| 2 | \(KE = \frac{1}{2}mv^2\) | Kinetic Energy (KE) of the projectile at the moment it leaves the spring. \( m \) is the mass and \( v \) is the velocity. |
| 3 | \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2\) | Applying conservation of energy, the EPE is converted to KE when the projectile leaves the spring. |
| 4 | \(kx^2 = mv^2\) | Eliminate the common factor (1/2) on both sides. |
| 5 | \(v = \sqrt{\frac{kx^2}{m}}\) | Solve for \( v \), the velocity of the projectile as it leaves the spring. |
| 6 | \(v = \sqrt{\frac{1191.67 \, \text{N/m} \cdot (0.120 \, \text{m})^2}{0.035 \, \text{kg}}}\) | Substitute the known values: \( k = 1191.67 \, \text{N/m} \), \( x = 0.120 \, \text{m} \), \( m = 0.035 \, \text{kg} \). |
| 7 | \(v = \sqrt{\frac{1191.67 \cdot 0.0144}{0.035}}\) | Calculate the values inside the equation. |
| 8 | \(v = \sqrt{489.67} \approx 22.12 \, \text{m/s}\) | Therefore, the speed of the projectile as it leaves the spring is approximately \( \boxed{22.12 \, \text{m/s}} \). |
Just ask: "Help me solve this problem."
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
A kickball is rolled by the pitcher at a speed of 10 m/s and it is kicked by another student. The kickball deforms a little during the kick, and then rebounds with a velocity of 15 m/s as its shape restores to a perfect sphere. Select all that must be true about the kickball and the kicking foot system.
A person holds a book at rest a few feet above a table. The person then lowers the book at a slow constant speed and places it on the table. Which of the following accurately describes the change in the total mechanical energy of the Earth–book system?
A spring launches a \(4 \, \text{kg}\) block across a frictionless horizontal surface. The block then ascends a \(30^\circ\) incline with a kinetic friction coefficient of \(\mu_k = 0.25\), stopping after \(55 \, \text{m}\) on the incline. If the spring constant is \(800 \, \text{N/m}\), find the initial compression of the spring. Disregard friction while in contact with the spring.
Two identical arrows, one with \( 2 \) times the speed of the other, are fired into a bale of hay. Assuming the hay exerts a constant “frictional” force on the arrows, the faster arrow will penetrate how much farther than the slower arrow?
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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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