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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. |
Just ask: "Help me solve this problem."
A spring stretches \( 8.0 \) \( \text{cm} \) when a \( 13 \) \( \text{N} \) force is applied. How far does it stretch when a \( 26 \) \( \text{N} \) force is applied?
A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?
A spring in a pogo-stick is compressed \( 12 \) \( \text{cm} \) when a \( 40. \) \( \text{kg} \) girl stands on it. What is the spring constant for the pogo-stick spring?
Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.
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.
\(\Delta x \approx 5.69\,\text{m}\)
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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] |
General Metric Conversion Chart
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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