Here is the solution to the problem posed:
(a) To find the force exerted by the motor, we start by using the formula that relates work, force, and distance:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(W = F \cdot d\) | Work done (\(W\)) is the product of the force (\(F\)) applied and the distance (\(d\)) over which the force is applied. |
| 2 | \(F = \frac{W}{d}\) | Rearrange the equation to solve for force. |
| 3 | \(F = \frac{520 \, \text{J}}{260 \, \text{m}}\) | Substitute the values for work (520 J) and distance (260 m) into the formula. |
| 4 | \(F = 2 \, \text{N}\) | Calculate the force exerted by the motor. |
(b) To determine the power generated by the motor, we use the definition of power as the rate of doing work:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(P = \frac{W}{t}\) | Power (\(P\)) is defined as the work done (\(W\)) per unit of time (\(t\)). |
| 2 | \(P = \frac{520 \, \text{J}}{37 \, \text{s}}\) | Substitute the values for work (520 J) and time (37 s). |
| 3 | \(P \approx 14.05 \, \text{W}\) | Calculate the power output of the motor. |
In (a), the force exerted by the motor is 2 N, and in (b), the power generated by the motor is 14.05 W.
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Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance \(d\). Ignore friction and assume that an equal force \(F\) is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
A box having a mass of \( 1.5 \) \( \text{kg} \) is accelerated across a table at \( 1.5 \) \( \text{m/s}^2 \). The coefficient of kinetic friction on the box is \( 0.3 \).
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving object?
Why is more fuel required for a spacecraft to travel from the Earth to the Moon than to return from the Moon to the Earth?
A ball is thrown straight up. At what point does the ball have the most energy?
A man weighing \( 700 \) \( \text{N} \) and a woman weighing \( 400 \) \( \text{N} \) have the same momentum. What is the ratio of the man’s kinetic energy \( K_m \) to that of the woman \( K_w \)?
A runner is moving at \( 4 \) \( \text{m/s} \). She is opposed by magic in the form of air resistance, which exerts a constant \( 20 \) \( \text{Newtons} \) force in the direction opposite her velocity. At what rate is she using energy to remain at constant velocity?
A small block of mass \( M \) is released from rest at the top of the curved frictionless ramp shown above. The block slides down the ramp and is moving with a speed \( 3.5v_0 \) when it collides with a larger block of mass \( 1.5M \) at rest at the bottom of the incline. The larger block moves to the right at a speed \( 2v_0 \) immediately after the collision.
Express your answers to the following questions in terms of the given quantities and fundamental constants.
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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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