Solving as a single conservation of energy equation:
| Derivation/Formula | Reasoning |
|---|---|
| \[W_F = F(6)\] | Work done by the applied horizontal force over \(6\,\text{m}\). |
| \[W_{f,h} = \mu_h m g (6)\] | Magnitude of work done by kinetic friction on the horizontal; friction is \(\mu_h m g\) and acts over \(6\,\text{m}\). |
| \[N_i = m g \cos\theta\] | Normal force on the incline is reduced by the angle, giving \(N_i\). |
| \[W_{f,i} = \mu_i m g \cos\theta\, \Delta x\] | Magnitude of work done by kinetic friction on the incline over distance \(\Delta x\). |
| \[U_g = m g\, \Delta x\, \sin\theta\] | Final gravitational potential energy; vertical rise is \(\Delta x\sin\theta\). Final kinetic energy is zero. |
| \[F(6) – \mu_h m g (6) – \mu_i m g \cos\theta\, \Delta x = m g\, \Delta x\, \sin\theta\] | Single energy balance: input work from the horizontal force minus both friction works equals the final gravitational potential energy. |
| \[F(6) – \mu_h m g (6) = m g\,(\sin\theta + \mu_i \cos\theta)\, \Delta x\] | Collect the \(\Delta x\) terms on the right and factor. |
| \[\Delta x = \frac{F(6) – \mu_h m g (6)}{m g\,(\sin\theta + \mu_i \cos\theta)}\] | Algebraic solution for \(\Delta x\). |
| \[\Delta x = \frac{110(6) – 0.25(12)(9.8)(6)}{(12)(9.8)\left(\sin(17^\circ) + 0.45\cos(17^\circ)\right)}\] | Substitute \(F = 110\,\text{N}\), \(\mu_h = 0.25\), \(\mu_i = 0.45\), \(m = 12\,\text{kg}\), \(g = 9.8\,\text{m/s}^2\), \(\theta = 17^\circ\). |
| \[\Delta x = \frac{660 – 176.4}{(12)(9.8)\left(\sin(17^\circ) + 0.45\cos(17^\circ)\right)}\] | Compute the numerator: \(110\times 6 = 660\), \(0.25\times 12\times 9.8\times 6 = 176.4\). |
| \[\sin(17^\circ) \approx 0.2924,\quad \cos(17^\circ) \approx 0.9563\] | Numerical trig values for \(17^\circ\). |
| \[\sin(17^\circ) + 0.45\cos(17^\circ) \approx 0.7227\] | Combine the angle terms in the denominator. |
| \[(12)(9.8)(0.7227) \approx 84.99\] | Evaluate \(m g(\sin\theta + \mu_i\cos\theta)\). |
| \[\Delta x \approx \frac{483.6}{84.99} \approx 5.7\,\text{m}\] | Final numerical evaluation for \(\Delta x\). |
| \[\boxed{\Delta x \approx 5.7\,\text{m}}\] | Distance slid up the incline before stopping. |
Alternatively you can split the motion up into (1) horizontal motion and (2) motion up the incline, then apply conservation of energy to each part to yield the same answer:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[N = m g\] | The normal force on a horizontal surface equals the object’s weight \(m g\). |
| 2 | \[f_k = \mu_k N = \mu_k m g\] | Kinetic friction magnitude is the product of coefficient \(\mu_k\) and normal force. |
| 3 | \[F_{\text{net}} = F_{\text{app}} – f_k\] | Net force equals applied force minus friction (opposite direction). |
| 4 | \[W_{\text{net}} = F_{\text{net}}\, \Delta x\] | Work by the net force over displacement \(\Delta x = 6\,\text{m}\). |
| 5 | \[W_{\text{net}} = \tfrac12 m v_x^2 – \tfrac12 m v_i^2\] | Work–energy theorem; the object starts from rest so \(v_i = 0\). |
| 6 | \[v_x = \sqrt{\frac{2 F_{\text{net}}\, \Delta x}{m}}\] | Solving the work–energy relation for the final speed. |
| 7 | \[v_x = \sqrt{\frac{2 (110\,\text{N} – 0.25\, (12\,\text{kg})(9.8\,\text{m/s}^2)) (6\,\text{m})}{12\,\text{kg}}} \;\approx\; 9.0\,\text{m/s}\] | Numeric substitution gives the speed at the base of the incline. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[KE_{\text{base}} = \tfrac12 m v_x^2\] | Kinetic energy as the object reaches the incline. |
| 2 | \[\Delta PE = m g (\Delta x \sin \theta)\] | Gravitational potential gain on an incline: height is \(\Delta x \sin \theta\). |
| 3 | \[W_{f} = -\mu_k m g \cos \theta \; \Delta x\] | Work done by kinetic friction along the incline (opposite motion). |
| 4 | \[KE_{\text{base}} = \Delta PE + |W_{f}|\] | All initial kinetic energy is dissipated by gravity and friction until rest. |
| 5 | \[\tfrac12 m v_x^2 = m g (\sin \theta + \mu_k \cos \theta) \, \Delta x\] | Combine energy losses (gravity + friction) into a single factor. |
| 6 | \[\Delta x = \frac{\tfrac12 m v_x^2}{m g (\sin \theta + \mu_k \cos \theta)}\] | Algebraic isolation of the distance up the incline. |
| 7 | \[\Delta x = \frac{\tfrac12 (12)(9.0^2)}{(12)(9.8)(\sin 17^\circ + 0.45 \cos 17^\circ)} \;\approx\; \boxed{5.7\,\text{m}}\] | Substituting numbers (\(\sin 17^\circ \approx 0.292\), \(\cos 17^\circ \approx 0.956\)) yields the sliding distance. |
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A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
A roller coaster ride at an amusement park lifts a car of mass \( 700 \, \text{kg} \) to point \( A \) at a height of \( 90 \, \text{m} \) above the lowest point on the track, as shown above. The car starts from rest at \( A \), rolls with negligible friction down the incline and follows the track around a loop of radius \( 20 \, \text{m} \). Point \( B \), the highest point on the loop, is at a height of \( 50 \, \text{m} \) above the lowest point on the track.
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance \(h\) above the floor. A block of mass \(M\) is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance \(x\). The block is released and strikes the floor a horizontal distance \(D\) from the edge of the table. Air resistance is negligible. Derive expressions for the following quantities only in terms of \(M, x, D, h,\) and any constants.
A block is attached to a horizontal spring and is initially at rest at the equilibrium position \( x = 0 \), as shown in Figure \( 1 \). The block is then moved to position \( x = -A \), as shown in Figure \( 2 \), and released from rest, undergoing simple harmonic motion. At the instant the block reaches position \( x = +A \), another identical block is dropped onto and sticks to the block, as shown in Figure \( 3 \). The two–block–spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy \( E_{\text{tot},2} \) of the two–block–spring system after the collision to the total mechanical energy \( E_{\text{tot},1} \) of the one–block–spring system before the collision?
A spring with spring constant \( k = 2.3 \) \( \text{N/m} \) is attached to an object of mass \( 10 \) \( \text{kg} \). If the object is hung from the ceiling by this spring, how much would the spring be stretched?
If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).
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?
\(\Delta x \approx 5.69\,\text{m}\)
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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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