AP Physics Unit

Unit 2 - Linear Forces




Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle \theta to the horizontal. If the coefficients of friction are µA = .2 and µB = .3 and if mA = mB = 5.0 kg determine:

  1. (a) The acceleration of the blocks. (3 points)
  2. (b) The tension in the cord, for an when the angle \theta = 32°. (3 points)

Objective: Determine the acceleration of the blocks and the tension in the cord given the masses, coefficients of friction, and incline angle given:

  • m_A = m_B = 5.0 , \text{kg}
  • \mu_A = 0.20
  • \mu_B = 0.30
  • \theta = 32^\circ
  • g = 9.8 , \text{m/s}^2 (acceleration due to gravity)

Part a: Calculate the acceleration of the blocks

Step Formula Derivation Reasoning
1 F_{\text{gravity, parallel A}} = m_A g \sin(\theta) Parallel component of gravitational force for A.
2 F_{\text{gravity, parallel B}} = m_B g \sin(\theta) Parallel component of gravitational force for B.
3 F_{\text{friction A}} = \mu_A m_A g \cos(\theta) Frictional force on A.
4 F_{\text{friction B}} = \mu_B m_B g \cos(\theta) Frictional force on B.
5 F_{\text{net A}} = F_{\text{gravity, parallel A}} – F_{\text{friction A}} Net force on A.
6 F_{\text{net B}} = F_{\text{gravity, parallel B}} – F_{\text{friction B}} Net force on B.
7 F_{\text{net}} = F_{\text{net B}} – F_{\text{net A}} Total net force on the system.
8 a = \frac{F_{\text{net}}}{m_A + m_B} Acceleration of the system.

Plug in the given values:

Step Formula Derivation Reasoning
9 a = \frac{(m_B g \sin(\theta) – \mu_B m_B g \cos(\theta)) – (m_A g \sin(\theta) – \mu_A m_A g \cos(\theta))}{m_A + m_B} Substitute the net forces from steps 5 and 6.
10 a = \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{5 + 5} Substitute given values.
11 a = \frac{(5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.30 \cdot \cos(32^\circ))) – (5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.20 \cdot \cos(32^\circ)))}{10} Simplify the expression.
12 a = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} Combine like terms.
13 a = \frac{9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} Simplify further.

Part b: Calculate the tension in the cord

Step Formula Derivation Reasoning
1 T = m_A \cdot a + F_{\text{friction A}} Tension equals the force to accelerate block A plus frictional force on A.
2 F_{\text{friction A}} = \mu_A \cdot m_A \cdot g \cdot \cos(\theta) Frictional force opposing the motion of block A.
3 T = m_A \cdot a + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) Substitute the frictional force into the tension formula.
4 T = m_A \cdot \left( \frac{F_{\text{net}}}{m_A + m_B} \right) + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) Substitute the expression for from the acceleration calculation.
5 T = 5 \cdot \left( \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) Insert given values for masses, coefficients of friction, gravitational acceleration, and angle.
6 T = 5 \cdot \left( \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) Simplify the expression for the net force component.
7 T = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) Further simplify the tension formula.
8 T = \frac{5 \cdot 9.8 \cdot 0.10 \cdot \cos(32^\circ)}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) Combine like terms for the final tension calculation.
9 T \approx 17.66 , \text{N} Calculate the numeric value for tension.
\boxed{T = 17.66 , \text{N}}
  1. .42 m/s2
  2. 17.66

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  1. .42 m/s2
  2. 17.66

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\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: \text{5 km}

  2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

  3. Perform the multiplication: \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}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}



Power of Ten




















(Base unit)


Deca- or Deka-


















  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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