Forces and Free-Body Diagrams Practice

Sample questions

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1. Sample 1difficulty 2/5

   Two forces act on a $4~\text{kg}$ object: $6~\text{N}$ east and $8~\text{N}$ north. What is the magnitude of the object's acceleration?

   • A

     $5.0~\text{m/s}^2$

   • B

     $0.5~\text{m/s}^2$

   • C

     $2.5~\text{m/s}^2$

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   • D

     $3.5~\text{m/s}^2$

   Why

   Net force $|\vec F| = \sqrt{6^2 + 8^2} = 10~\text{N}$. $a = F/m = 10/4 = 2.5~\text{m/s}^2$.

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2. Sample 2difficulty 2/5

   A $2.0~\text{kg}$ mass hangs in equilibrium from a vertical spring ($k = 200~\text{N/m}$). Using $g \approx 10~\text{m/s}^2$, by how much is the spring stretched from its natural length?

   • A

     $0.05~\text{m}$

   • B

     $0.10~\text{m}$

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   • C

     $0.20~\text{m}$

   • D

     $0.01~\text{m}$

   Why

   $k x = m g \Rightarrow x = m g / k = (2 \cdot 10)/200 = 0.10~\text{m}$.

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3. Sample 3difficulty 2/5

   The diagram shows the FBD of a person on a scale in an elevator. If the elevator accelerates <strong>upward</strong>, then

   • A

     $N > W$

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   • B

     $N = 0$

   • C

     $N < W$

   • D

     $N = W$

   Why

   $N - W = m a$. Upward $a > 0$ → $N > W$.

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4. Sample 4difficulty 2/5

   A scale shows zero weight for a person standing on it inside an elevator. The elevator must be

   • A

     At rest.

   • B

     Moving up at constant speed.

   • C

     Moving down at constant speed.

   • D

     In free fall.

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   Why

   Zero scale reading means $N = 0$, so $a = -g$ — the elevator (and person) is in free fall.

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5. Sample 5difficulty 2/5

   The free-body diagram shows weight ($W$), normal ($N$), and friction ($f$) on a block sliding <strong>down</strong> a rough incline. Friction here points

   • A

     Vertically down

   • B

     Up the slope

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   • C

     Toward the bottom of the incline

   • D

     Down the slope

   Why

   Kinetic friction opposes motion. The block slides down, so $f$ points up the slope.

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