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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 blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance \(d\). Ignore friction and assume that an equal force \(F\) is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?
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?
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?
You are working out on a rowing machine. Each time you pull the rowing bar toward you, it moves a distance of \(1.25 \, \text{m}\) in a time of \(0.84 \, \text{s}\). The readout on the display indicates that the average power you are producing is \(76 \, \text{W}\). What is the magnitude of the force that you exert on the handle?
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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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