| Step | Reasoning |
|---|---|
| Identify the relationship between external work and system energy. \[ W_{ext} = \Delta E_{sys} \] |
The question asks to evaluate the energy change of a system based on its definition. |
| Analyze the ‘sled-only’ system. \[ W_{ext} = W_{pull} + W_{f} + W_N + W_g \] |
To evaluate choices A and B, we must identify all external forces acting on the sled. |
| Calculate the net work for the sled-only system. \[ \begin{align*} \Sigma F_x &= F_{pull} – f_k = 0 \implies F_{pull} = f_k \\ W_{ext} &= (F_{pull})(d) + (-f_k)(d) + 0 + 0 \\ W_{ext} &= (F_{pull} – f_k)d = 0 \end{align*} \] |
Since the velocity is constant, the net force is zero, allowing us to relate the pulling force to the friction force. |
| Determine the change in energy for the sled-only system. \[ \Delta E_{sled} = W_{ext} = 0 \] |
The work-energy theorem connects the calculated net work to the system’s energy change. |
| Analyze the ‘sled-and-surface’ system to verify the remaining choices. \[ W_{ext} = W_{pull} \] |
Changing system boundaries changes which forces are external and which are internal. |
| Conclusion for the combined system. \[ \Delta E_{sys} = W_{pull} = F_{pull}d > 0 \] |
In the combined system, friction is internal and converts mechanical energy to thermal energy, but the pulling force is still external and does positive work. |
Why each choice is correct or incorrect:
(A) Fails to account for the negative work done by friction, which is an external force in the sled-only system.
(B) This is the correct answer; the net external work on the sled is zero because the positive work of the pull cancels the negative work of friction.
(C) Misidentifies the system as isolated; the pulling force is external and adds energy to the system, causing the total energy to increase.
(D) Incorrectly states total energy decreases; while thermal energy is generated, the external pulling force does positive work that increases the total energy of the system.
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Students are investigating the periodic motion of a physical pendulum to determine the acceleration due to gravity \(g\). They are provided with a uniform rigid rod of length \(L = 1.20 \text{ m}\) and mass \(M\). The rod has several small holes drilled along its length, allowing it to be pivoted about a frictionless axle. The distance from the center of mass of the rod to the pivot is \(x\).
When displaced by a small angle \(\theta\) and released, the rod oscillates. The theoretical period of oscillation \(T\) is given by the equation:
\[ T = 2\pi \sqrt{ \dfrac{\dfrac{1}{12}L^2 + x^2}{gx} } \]

A group of students is investigating the properties of a horizontal spring system. A cart of known mass \(M_c\) is attached to an ideal horizontal spring of unknown spring constant \(k\). The other end of the spring is attached to a fixed rigid wall. The cart rolls on a horizontal track with negligible friction. A motion sensor is positioned at the far end of the track to measure the speed of the cart. The students are provided with a set of identical metal blocks, each of known mass \(m_0\), which can be securely attached to the top of the cart. The students can pull the cart to stretch the spring by a distance \(D\) from its equilibrium position before releasing it from rest.

A solid cylinder of mass \(M\) and radius \(R\) is released from rest at the top of an incline of length \(L\) that makes an angle \(\theta\) with the horizontal, as shown in Figure 1. The cylinder rolls down the incline without slipping. As the cylinder moves, the air exerts a constant drag force of magnitude \(F_D\) on the cylinder, directed opposite to its velocity. The drag force acts entirely at the center of mass of the cylinder. The rotational inertia of a solid cylinder is \(I = \dfrac{1}{2}MR^2\).

A uniform rigid rod of mass \( M \) and length \( L \) hangs at rest from a fixed, frictionless pivot at its top end. A small block of mass \( m \) is sliding horizontally with speed \( v_0 \) toward the rod. The block strikes the rod at a distance \( d \) below the pivot and sticks to it. The rotational inertia of the rod about the pivot is \( \dfrac{1}{3}ML^2 \).

A uniform horizontal shelf of mass M is attached to a vertical wall by a hinge at its left end and is supported by a vertical wire at its right end. A student intends to use the rotational equilibrium condition \(\sum \tau = 0\) to determine the magnitude of the tension in the wire. The student knows the mass of the shelf but does not know the magnitude or direction of the force exerted by the hinge on the shelf. Which of the following describes the most useful location for the axis of rotation and provides a correct justification?

Block \(A\) of mass \(m_A\) rests on a rough horizontal table. Block \(B\) of mass \(m_B\) rests on top of Block \(A\). A light, inextensible string is attached to Block \(A\), passes over an ideal pulley at the edge of the table, and is attached to Block \(C\) of mass \(m_C\), which hangs vertically, as shown in Figure 1.
The coefficient of kinetic friction between Block \(A\) and the table is \(\mu_k\). The coefficient of static friction between Block \(A\) and Block \(B\) is \(\mu_s\). The system is released from rest, and Block \(B\) moves along with Block \(A\) without slipping. Assume air resistance is negligible.

Block A of mass \(m_A\) is traveling at speed \(v_0\) to the right on a horizontal, frictionless surface. An ideal, massless spring of spring constant \(k\) is attached to the right face of Block A. Block B of mass \(m_B\) is initially at rest on the surface. Block A collides with Block B, compressing the spring.

A solid sphere is launched across a rough horizontal surface with an initial translational velocity \(v_0\) to the right and zero initial angular velocity. As the sphere moves, the frictional force from the surface causes the sphere’s translational speed to decrease and its angular speed to increase until it eventually rolls without slipping. Which of the following correctly identifies the type and direction of the frictional force exerted by the surface on the sphere while it is sliding before it begins to roll without slipping?

A picture frame of mass Equivalence mass is suspended from a small, frictionless peg by a light, non-stretchable wire of total length \(L\). The ends of the wire are attached to two points at the top corners of the frame that are a distance \(w\) apart. The peg is positioned such that the two wire segments are of equal length and the system is in static equilibrium.
Which of the following expressions correctly represents the tension in the wire?

Two different spring-mass systems, A and B, oscillate in simple harmonic motion on a horizontal, frictionless surface. The potential energy \(U\) of each system as a function of displacement \(x\) from equilibrium is shown in the graph. The mass of the block in system B is twice the mass of the block in system A (\(m_B = 2m_A\)). What is the ratio \(T_B / T_A\) of the periods of the two systems?
B
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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