| 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 \( \boxed{22.12 \, \text{m/s}} \). |
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A \( 7.3 \) \( \text{kg} \) mass is placed on a spring with a spring constant of \( 34 \) \( \text{N/cm} \). How much does this stretch the spring?
A box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
Two blocks, \( m_2 > m_1 \), having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient \( \mu_k \). Which block will travel further before stopping.
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
A person is making homemade ice cream. She exerts a force of magnitude \(23 \, \text{N}\) on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius \(0.25 \, \text{m}\). The force is always applied parallel to the motion of the handle. If the handle is turned once every \(1.7 \, \text{s}\), what is the average power being expended?
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
An object at rest suddenly explodes into two fragments (\(m_1\) and \(m_2\)) by an explosion. Fragment \(m_1\) acquires \(3\) times the kinetic energy of the other. What is the ratio of \(m_1\) to \(m_2\)?
A 0.4 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.2 J. The potential energy as a function of position presented by the graph.
A ski lift carries skiers along a \(695 \, \text{m}\) slope inclined at \(34^\circ\). To lift a single rider, it is necessary to move \(72 \, \text{kg}\) of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If \(65\%\) of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
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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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