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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 \( \boxed{22.12 \, \text{m/s}} \). |
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If a small motor does 520 J of work to move a toy car 260 meters in a time of 37 seconds.
A typical \( 68 \text{-kg} \) person generates a steady mechanical power output of \( 120 \text{ W} \) at the pedals of a bicycle. Approximately how many Calories are “burned” (total metabolic energy expended) when the person rides a bicycle for \( 15 \text{ minutes} \)? A typical energy efficiency for the human body is \( 25\% \), which takes into account the release of thermal energy. Note (\( 1 \text{ Cal} = 4186 \text{ J} \)).
A big bird has a mass of about 0.021 kg. Suppose it does 0.36 J of work against gravity, so that it ascends straight up with a net acceleration of 0.625 m/s2. How far up does it move?
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
Refer to the diagram above and solve all equations in terms of \(R\), \(M\), \(k\), and constants.
A \(81 \, \text{kg}\) student dives off a \(45 \, \text{m}\) tall bridge with an \(18 \, \text{m}\) long bungee cord tied to his feet and to the bridge. You can consider the bungee cord to be a flexible spring. What spring constant must the bungee cord have for the student’s lowest point to be \(2.0 \, \text{m}\) above the water?
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
What force is necessary to stretch an ideal spring with a spring constant of \( 120 \) \( \text{N/m} \) by \( 30 \) \( \text{cm} \)?
How does the speed \(v_1\) of a block \(m\) reaching the bottom of slide 1 compare with \(v_2\), the speed of a block \(2m\) reaching the end of slide 2? The blocks are released from the same height.
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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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