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 block of mass \( 0.5 \) \( \text{kg} \) is attached to a horizontal spring with a spring constant of \( 150 \) \( \text{N/m} \). The block is released from rest at position \( x = 0.05 \) \( \text{m} \), as shown, and undergoes simple harmonic motion, reaching a maximum position of \( x = 0.1 \) \( \text{m} \). The speed of the block when it passes through position \( x = 0.09 \) \( \text{m} \) is most nearly
A particle of mass \(m\) slides down a fixed, frictionless sphere of radius \(R\), starting from rest at the top.
In terms of \(m\), \(g\), \(R\), and \(\theta\), determine each of the following for the particle while it is sliding on the sphere.
Two masses \(m_1\) and \(4m_1\) are on an incline. Both surfaces have the same coefficient of kinetic friction. Both objects start from rest at the same height. Which mass has the largest speed at the bottom?
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?
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 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 stone is falling at a constant velocity vertically down a tube filled with oil. Which of the following statements about the energy changes of the stone during its motion are correct?
I. The gain in kinetic energy is less than the loss in gravitational potential energy.
II. The sum of kinetic and gravitational potential energy of the stone is constant.
III. The work done by the force of gravity has the same magnitude as the work done by friction.
A child pushes horizontally on a box of mass m with constant speed v across a rough horizontal floor. The coefficient of friction between the box and the floor is µ. At what rate does the child do work on the box?
An object with a mass \(m = 80 \, \text{g}\) is attached to a spring with a force constant \(k = 25 \, \text{N/m}\). The spring is stretched \(52.0 \, \text{cm}\) and released from rest. If it is oscillating on a horizontal frictionless surface, determine the velocity of the mass when it is halfway to the equilibrium position.
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?
\(\Delta x \approx 5.69\,\text{m}\)
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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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