AP Physics 1 Force and Translational Dynamics — Worked Answer Explanations

Unit 2 · 18% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Physics 1 Force and Translational Dynamics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Force and Translational Dynamics practice test and come back here to review, or head back to the Force and Translational Dynamics unit overview.

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Question 1 · Easy
A $5.0 kg$ block on a frictionless horizontal surface is pulled by a horizontal force of $20 N$ . What is the block's acceleration?
- A
  $0.25 m/s^{2}$
  Why not A: Inverted Newton's second law, divided mass by force.
- B
  $4.0 m/s^{2}$ Correct
- C
  $15 m/s^{2}$
  Why not C: Subtracted mass from force.
- D
  $100 m/s^{2}$
  Why not D: Multiplied force by mass.
Explanation
Apply Newton's second law: $a = F / m = 20/5.0 = 4.0 m/s^{2}$ .

Key takeaway
$F_{net} = ma$ — acceleration scales linearly with net force and inversely with mass.
Question 2 · Easy
A $2.0 kg$ block sits on a horizontal surface with coefficient of static friction $μ_{s} = 0.40$ . Taking $g = 10 m/s^{2}$ , what is the maximum horizontal force that can be applied without the block sliding?
- A
  $0.80 N$
  Why not A: Forgot to multiply by $g$ .
- B
  $5.0 N$
  Why not B: Divided by $μ_{s}$ instead of multiplying.
- C
  $8.0 N$ Correct
- D
  $20 N$
  Why not D: Used the weight without applying $μ_{s}$ .
Explanation
Maximum static friction is $f_{s} = μ_{s} N = μ_{s} m g = 0.40 \times 2.0 \times 10 = 8.0 N$ . Any applied force at or below this magnitude keeps the block stationary.

Key takeaway
Static friction provides whatever force is needed to prevent motion, up to a maximum of $\mu_s N$.
Question 3 · Easy
A $4.0 kg$ box experiences three horizontal forces: $10 N$ east, $6.0 N$ west, and $4.0 N$ east. What is the box's acceleration?
- A
  $1.0 m/s^{2}$ east
  Why not A: Mis-summed the forces.
- B
  $2.0 m/s^{2}$ eastCorrect
- C
  $5.0 m/s^{2}$ east
  Why not C: Forgot to subtract the westward force.
- D
  $8.0 m/s^{2}$ east
  Why not D: Summed magnitudes without considering direction.
Explanation
Net force: $10 + 4 - 6 = 8 N$ east. $a = F_{n e t} / m = 8/4 = 2.0 m/s^{2}$ east.

Key takeaway
Sum forces vectorially (with sign for direction) before applying $F = ma$.
Question 4 · Medium
Two blocks of masses $3.0 kg$ and $5.0 kg$ are in contact on a frictionless surface. A horizontal force of $24 N$ is applied to the $3.0 kg$ block, pushing both. What is the contact force between the blocks?
- A
  $9.0 N$
  Why not A: Used the smaller mass alone with the system acceleration.
- B
  $15 N$ Correct
- C
  $24 N$
  Why not C: Confused contact force with applied force.
- D
  $40 N$
  Why not D: Added masses incorrectly.
Explanation
First find system acceleration: $a = 24/ (3 + 5) = 3.0 m/s^{2}$ . The contact force on the $5.0 kg$ block must alone produce this acceleration: $F_{co n t a c t} = m_{2} a = 5.0 \times 3.0 = 15 N$ .

Key takeaway
Find system acceleration first, then isolate one body to compute internal contact forces.
Question 5 · Medium
An object hangs from a spring scale inside an elevator. When the elevator is at rest the scale reads $50 N$ . When the elevator accelerates upward at $2.0 m/s^{2}$ (with $g = 10 m/s^{2}$ ), what does the scale read?
- A
  $10 N$
  Why not A: Subtracted weight and acceleration directly without considering scale reads tension.
- B
  $40 N$
  Why not B: Subtracted instead of added the inertial term.
- C
  $60 N$ Correct
- D
  $100 N$
  Why not D: Doubled the rest reading.
Explanation
The scale reads the tension. With mass $m = 50/10 = 5.0 kg$ and upward acceleration $a$ : $T - m g = ma$ , so $T = m (g + a) = 5.0 \times 12 = 60 N$ .

Key takeaway
Apparent weight in an accelerating frame is $m(g \pm a)$; upward acceleration increases the scale reading.
Question 6 · Medium
A block slides at constant velocity down an inclined plane making angle $θ$ with the horizontal. What is the coefficient of kinetic friction between the block and the surface?
- A
  $sin θ$
  Why not A: Forgot to divide by the cosine for the normal direction.
- B
  $cos θ$
  Why not B: Confused the parallel and perpendicular weight components.
- C
  $tan θ$ Correct
- D
  $\frac{1}{tan θ}$
  Why not D: Inverted sine and cosine.
Explanation
Constant velocity means $F_{n e t} = 0$ . Down-slope: $m g sin θ = μ_{k} N = μ_{k} m g cos θ$ . Solving: $μ_{k} = tan θ$ .

Key takeaway
On an incline at terminal angle, $\mu_k = \tan\theta$.
Question 7 · Medium
Two masses, $m_{1} = 2.0 kg$ and $m_{2} = 3.0 kg$ , are connected by a massless string over a frictionless pulley (Atwood machine). With $g = 10 m/s^{2}$ , what is the magnitude of acceleration of the system?
- A
  $1.0 m/s^{2}$
  Why not A: Used average mass instead of total.
- B
  $2.0 m/s^{2}$ Correct
- C
  $5.0 m/s^{2}$
  Why not C: Used the heavier mass's weight divided by total mass.
- D
  $10 m/s^{2}$
  Why not D: Confused with free-fall acceleration.
Explanation
Net force on system is $(m_{2} - m_{1}) g$ , total mass is $m_{1} + m_{2}$ : $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}} = \frac{1 \cdot 10}{5} = 2.0 m/s^{2}$ .

Key takeaway
Atwood acceleration is $\dfrac{|m_2-m_1| g}{m_1+m_2}$.
Question 8 · Hard
A $1500 kg$ car rounds a flat (unbanked) curve of radius $50 m$ at $15 m/s$ . What is the minimum coefficient of friction needed to keep the car on the curve? (Take $g = 10 m/s^{2}$ .)
- A
  $0.15$
  Why not A: Off by an order of magnitude in the centripetal calculation.
- B
  $0.30$
  Why not B: Used $v$ instead of $v^{2}$ .
- C
  $0.45$ Correct
- D
  $0.90$
  Why not D: Doubled the answer, perhaps by squaring twice.
Explanation
Friction provides centripetal force: $μ m g \geq m v^{2} / r$ , so $μ \geq v^{2} / (r g) = 225/ (50 \times 10) = 0.45$ .

Key takeaway
On a flat curve, $\mu_{min} = v^2/(rg)$ — independent of mass.