A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).(a) i. Sketch and label a graph of the block’s kinetic energy \( K \) versus position \( x \) for \( -D \le x \le 4D \), using an appropriate common vertical scale. ii. Sketch and label a graph of the gravitational potential energy \( U_g \) of the block–Earth system versus position \( x \) for \( -D \le x \le 4D \), using the same vertical scale as in the previous graph. , (b) The block is released from the top of a new ramp that has a base length of \(2D\), but still makes an angle \(\theta\) with the horizontal. A student is asked to predict whether the final horizontal position of the block will be twice as far from the base of the ramp compared to when it was released from the original ramp. The student reasons that since the block will be released from a new height that is twice as high as the original height, the block will have more energy when it reaches the base of the ramp, so it will slide farther along the right surface before stopping at a new position. 1. State the aspects of the student’s reasoning that are correct, or write “none” if no aspects are correct. 2. State the aspects of the student’s reasoning that are incorrect, or write “none” if no aspects are incorrect. , (c) Derive an expression for the block’s new final horizontal position after sliding from the taller ramp, expressing your answer in terms of \( D \). , (d) Explain how the mathematical relationships from part (e) justify any correct aspects you listed and correct any incorrect aspects you listed in parts (b)(i) and (b)(ii). :

A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).(a) On the axes provided, sketch and label graphs of the following quantities as a function of the position \( x \) of the block for \( -D \le x \le 4D \). Both graphs must use the same vertical scale. i. The kinetic energy \( K \) of the block ii. The gravitational potential energy \( U_g \) of the block–Earth system(b) The block is now released from the top of a different ramp that still makes the same angle \( \theta_0 \) with the horizontal but whose base length is \( 2D \). A student is asked whether the block’s final horizontal position will now be twice as far (i.e., at \( x = 8D \)) compared with the original situation. The student reasons that, because the new height is twice the original height, the block will have more energy at the base of the new ramp and therefore will slide farther along the horizontal surface until stopping at \( x = 8D \). i. Which aspects of the student’s reasoning, if any, are correct? If none are correct, write “none”. ii. Which aspects of the student’s reasoning, if any, are incorrect? If none are incorrect, write “none”.(c) Derive an equation for the new final position of the block in terms of \( D \).(d) Referring to the mathematical relationships you obtained in part (c): • For any correct aspects identified in part (b)(i), explain how your relationships support the student’s reasoning. • For any incorrect aspects identified in part (b)(ii), explain how your relationships correct the student’s reasoning. • Nerd Notes

AP Physics

Unit 4 - Energy

FRQ

Mathematical

Intermediate

Part (a) – Energy graphs

Derivation or Formula	Reasoning
\[U_g(x)=mg(-x)\tan\theta_0 \quad ( -D\le x\le 0)\]	Height above the horizontal surface is \((-x)\tan\theta_0\); choose zero gravitational energy at the base (\(x=0\)). Thus the potential energy on the ramp varies linearly with the horizontal coordinate.
\[K(x)=mgD\tan\theta_0-mg(-x)\tan\theta_0=mg\tan\theta_0(D+x) \quad ( -D\le x\le 0)\]	Mechanical energy is conserved on the friction-free ramp, so kinetic energy equals the difference between the initial potential energy \(mgD\tan\theta_0\) and the remaining potential energy \(U_g(x)\).
\[U_g(x)=0 \quad (0\le x\le 4D)\]	On the horizontal surface the height is zero; gravitational potential energy remains constant at the chosen zero reference.
\[K(x)=mgD\tan\theta_0-\mu_k mg\,x \quad (0\le x\le 4D)\]	Friction does work \(W_f=-\mu_k mg\,x\). Kinetic energy therefore decreases linearly from its value at the base to zero at \(x=4D\).

Part (b) – Evaluate student’s reasoning

Derivation or Formula	Reasoning
\[\text{–}\]	Correct aspects: Doubling the ramp height doubles both the gravitational potential energy at the top and the kinetic energy at the base, so a greater distance is expected on the rough surface. The prediction of a final position at \(x=8D\) is therefore consistent. Incorrect aspects: none.

Derivation or Formula

Reasoning

\[\text{–}\]

Correct aspects: Doubling the ramp height doubles both the gravitational potential energy at the top and the kinetic energy at the base, so a greater distance is expected on the rough surface. The prediction of a final position at \(x=8D\) is therefore consistent.

Incorrect aspects: none.

Part (c) – New stopping position

Derivation or Formula	Reasoning
\[h=D\tan\theta_0\]	Initial ramp height in the first situation, expressed with base length \(D\).
\[mg h = \mu_k mg(4D)\]	Work–energy theorem: kinetic energy at the base equals the work done by friction over \(4D\), allowing \(\mu_k\) to be related to \(\theta_0\).
\[\mu_k = \frac{\tan\theta_0}{4}\]	Solve the previous relationship for \(\mu_k\).
\[h’ = 2D\tan\theta_0\]	The new ramp is twice as long horizontally, so its height is doubled.
\[mg h’ = \mu_k mg\,x_f\]	Set the new kinetic energy at the base equal to the work done by friction over the unknown stopping distance \(x_f\).
\[x_f = \frac{2D\tan\theta_0}{\mu_k} = \frac{2D\tan\theta_0}{\tan\theta_0/4}=8D\]	Substitute \(\mu_k\) from above to obtain the new final position.
\[\boxed{x_f = 8D}\]	Final algebraic result: the block stops at \(x=8D\) relative to the base of the ramp.

Part (d) – Connect to student’s reasoning

Derivation or Formula	Reasoning
\[K_{\text{base}}\propto h\]	The equations show that kinetic energy at the base is directly proportional to the ramp height; doubling \(h\) indeed doubles \(K_{\text{base}}\), validating that part of the student’s claim.
\[x_f = \frac{K_{\text{base}}}{\mu_k mg}\]	Since the friction force is unchanged, the stopping distance is proportional to \(K_{\text{base}}\). Because \(K_{\text{base}}\) doubles, the horizontal distance also doubles, fully supporting the student’s predicted position \(x=8D\).
\[\text{–}\]	No steps contradict the student’s logic, so there are no incorrect aspects to correct.

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Energy, Friction, Inclines

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1D Kinematics, Friction, Inclines, Linear Forces

\(x_f=8D\)

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]

Prefix	Symbol	Power of Ten
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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AP Physics

Unit 4 - Energy

FRQ

Mathematical

Intermediate

Pro

Pro

Educator

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Part (a) – Energy graphs

Part (b) – Evaluate student’s reasoning

Part (c) – New stopping position

Part (d) – Connect to student’s reasoning

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