Answer: C. Simple explanation: break each vector down into its \(x\) and \(y\) components. You’ll be able to see that all the vector components along both axis cancel out resulting in \(0\) net force (and therefore constant velocity).
| Option | Explanation |
|---|---|
| (a) | A free-body diagram that has unbalanced forces – the object would accelerate in the direction of the resultant force. |
| (b) | A free-body diagram that has unbalanced forces – similar to option (a), the object would accelerate rather than move at a constant velocity. |
| (c) | A free-body diagram that shows all forces balanced – the net force acting on the object is zero, which means the object could potentially move at a constant velocity. |
| (d) | A free-body diagram showing unbalanced forces – so similar to options (a) and (b), this would result in the object accelerating. |
In this situation where the object is moving at constant velocity, the net force acting on it must be zero, meaning all forces must be balanced. Therefore, the correct answer is option (c), which represents a situation where the forces in all directions are equal, allowing the object to maintain constant velocity.
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A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
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 the coefficient of static friction is \( \mu_s = 0.5 \), how much force must be applied to a spring (spring constant of \( 0.8 \) \( \text{N/m} \)) which is attached to a block of wood (mass \( 4.0 \) \( \text{kg} \)) in order to just begin to move the block?
A spring is connected to a wall and a horizontal force of \( 80.0 \) \( \text{N} \) is applied. It stretches \( 25 \) \( \text{cm} \); what is its spring constant?
A \(2.2 \times 10^{21} \, \text{kg}\) moon orbits a distant planet in a circular orbit of radius \(1.5 \times 10^8 \, \text{m}\). It experiences a \(1.1 \times 10^{19} \, \text{N}\) gravitational pull from the planet. What is the moon’s orbital period in Earth days?
Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the radius of a geosynchronous orbit about the neutron star? The mass of the sun can be found in the formula sheet above.
Two balls have their centers \( 2.0 \) \( \text{m} \) apart. One ball has a mass of \( 8.0 \) \( \text{kg} \). The other has a mass of \( 6.0 \) \( \text{kg} \). What is the gravitational force between them?
Two wires support an unknown mass as shown in the diagram. The tension in the left wire is measured to be \( 17.5 \) \( \text{N} \) and the tension in the right wire is \( 30.3 \) \( \text{N} \). The left wire makes an angle of \( 30^{\circ} \) with the horizontal, and the right wire makes an angle of \( 60^{\circ} \) with the horizontal. What is the mass of the object?
A ball of mass \( m \) is suspended from two strings of unequal length as shown above. The magnitudes of the tensions \( T_1 \) and \( T_2 \) in the strings must satisfy which of the following relations?
A ring of negligible mass remains in static equilibrium under the influence of three coplanar forces, as shown in the accompanying diagram. Two forces, each with a magnitude of \(10 \, \text{N}\), act on the ring at an angle \(\theta\) above the horizontal—one directed to the right and the other to the left. A third force, \(F\), acts vertically downward. Determine the magnitude of force \(F\).
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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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