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 \( 1.0 \)\( \text{-kg} \) object is moving with a velocity of \( 6.0 \) \( \text{m/s} \) to the right. It collides and sticks to a \( 2.0 \)\( \text{-kg} \) object moving with a velocity of \( 3.0 \) \( \text{m/s} \) in the same direction. How much kinetic energy was lost in the collision?
A \( 0.0350 \) \( \text{kg} \) bullet moving horizontally at \( 425 \) \( \text{m/s} \) embeds itself into an initially stationary \( 0.550 \) \( \text{kg} \) block.
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 uniform solid cylinder of mass \( M \) and radius \( R \) is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
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 cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
A boulder is raised above the ground so that its potential energy is \(550 \, \text{J}\). Then it is dropped. Assuming \(92 \, \text{J}\) of energy was lost to air resistance, what is the kinetic energy of the boulder just before it hits the ground?
An elastic cord is \( 80\) \( \text{cm} \) long when it is supporting a mass of \( 10. \) \( \text{kg} \) hanging from it at rest. When an additional \( 4.0 \) \( \text{kg} \) is added, the cord is \( 82.5 \) \( \text{cm} \) long.
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?
\(\Delta x \approx 5.69\,\text{m}\)
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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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