In the diagram shown a 20 N force is applied to a block B (7 kg). Block A has a mass of 3 kg. Assume frictionless conditions.(a) Calculate the acceleration of the system , (b) Calculate the normal force between the 2 masses , (c) Bonus: What would happen to the acceleration and normal force if the force was applied to block A instead? :

2 m/s2 6 N In this case applying the force to block A would not change the acceleration or the normal force between the two block. If the floor had friction this would not be the case.

In the diagram shown a 20 N force is applied to a 7 kg mass. Assume frictionless conditions. Calculate the acceleration of the system Calculate the normal force between the 2 masses Bonus: What would happen to the acceleration and normal force if the force was applied to block A instead • Nerd Notes

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AP Physics

Unit 2 - Linear Forces

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Mathematical

FRQ

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Part a: Calculate the acceleration of the system

Step	Formula Derivation	Reasoning
1	[katex] F_{\text{net}} = m_{\text{total}} \cdot a [/katex]	Net force equals total mass times acceleration.
2	[katex] a = \frac{F_{\text{net}}}{m_{\text{total}}} [/katex]	Solve for acceleration $a$ .

Let F_net = 20 N and total mass be (3 kg + 7 kg) = 10 kg.

Step	Formula Derivation	Reasoning
3	[katex] a = \frac{20}{3 + 7} [/katex]	Plug in the values for F_net and total mass.
4	[katex] a = \frac{20}{10} [/katex]	Simplify the denominator.
5	[katex] a = 2 , \text{m/s}^2 [/katex]	Calculate the acceleration.

[katex] \boxed{a = 2 , \text{m/s}^2} [/katex]

Part b: Calculate the normal force between the 2 masses

Under frictionless conditions, the normal force between the masses is equal to the force that the 7 kg mass would exert on the 3 kg mass due to the acceleration of the system.

Step	Formula Derivation	Reasoning
1	[katex] F_{\text{normal}} = m_{3\text{kg}} \cdot a [/katex]	Normal force equals mass of 3 kg times acceleration.
2	[katex] F_{\text{normal}} = 3 \cdot 2 [/katex]	Use the previously calculated acceleration.
3	[katex] F_{\text{normal}} = 6 , \text{N} [/katex]	Calculate the normal force.

[katex] \boxed{F_{\text{normal}} = 6 , \text{N}} [/katex]

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2 m/s2
6 N
In this case applying the force to block A would not change the acceleration or the normal force between the two block. If the floor had friction this would not be the case.

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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

Conversion Example

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
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]

1. Some answers may vary by 1% due to rounding.
Gravity values may differ: $9.81 \, \text{m/s}^2$ or $10 \, \text{m/s}^2$.
Variables can be written differently. For example, initial velocity ($v_i$) may be $u$, and displacement ($\Delta x$) may be $s$.
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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
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

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