AP Physics C: Mechanics Work, Energy, and Power — Worked Answer Explanations

Unit 3 · 12 questions explained

1. Question 1 · Easy

   A constant force $F=10$ N is applied at $0°$ to a block that moves 5 m along a frictionless surface. How much work is done by $F$?

   • A
     $0$ J

     Why not A: Thought forces perpendicular to motion do zero work — but $F$ is parallel here.
   • B
     $2$ J

     Why not B: Divided instead of multiplied.
   • C
     $50$ JCorrect
   • D
     $500$ J

     Why not D: Multiplied by $10^2$ instead of $10\times 5$.
   Explanation

   $W=F\cdot d\cdot \cos 0°=10\times 5\times 1=50$ J. Work is force times displacement when force and displacement are parallel.

   Key takeaway

   $W = \mathbf{F} \cdot \Delta\mathbf{r} = F\,d\cos\theta$. When $\theta = 0°$, all of $F$ contributes.

2. Question 2 · Easy

   A 3 kg object moves from rest and reaches a speed of $4$ m/s. What is the net work done on it?

   • A
     $12$ J

     Why not A: Used $W=mv$ instead of $\frac{1}{2}m{v}^2$.
   • B
     $24$ JCorrect
   • C
     $48$ J

     Why not C: Forgot the factor of $\frac{1}{2}$.
   • D
     $6$ J

     Why not D: Divided $\frac{1}{2}m{v}^2$ by $v$ rather than evaluating it.
   Explanation

   Work–energy theorem: $W_{net}=\Delta KE=\frac{1}{2}m{v}^2-0=\frac{1}{2}(3)(16)=24$ J.

   Key takeaway

   The net work done on an object equals the change in kinetic energy: $W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.

3. Question 3 · Easy

   A spring with constant $k=200$ N/m is compressed 0.1 m from its natural length. How much elastic potential energy is stored?

   • A
     $0.1$ J

     Why not A: Computed $kx$ instead of $\frac{1}{2}k{x}^2$.
   • B
     $1$ JCorrect
   • C
     $2$ J

     Why not C: Forgot the factor of $\frac{1}{2}$.
   • D
     $20$ J

     Why not D: Used $k\cdot x$ with $x=0.1$ and then multiplied by 10.
   Explanation

   Elastic potential energy stored in a spring compressed by $x$: $U_s=\frac{1}{2}k{x}^2$. With $k=200$ N/m and $x=0.1$ m: $U_s=\frac{1}{2}(200)(0.1{)}^2=\frac{1}{2}(200)(0.01)=1$ J. This energy is recovered when the spring returns to its natural length.

   Key takeaway

   Elastic potential energy is $U_s = \frac{1}{2}kx^2$ — note the square on $x$ and the factor of $\frac{1}{2}$. The formula comes from integrating the spring force $F = kx$ over displacement.

4. Question 4 · Easy

   A 2 kg block slides 4 m down a frictionless incline inclined at 30°. Using the work–energy theorem, what is the final kinetic energy? ($g=10$ m/s²)

   • A
     $20$ J

     Why not A: Used height $=d$ instead of $h=d\sin 30°$.
   • B
     $40$ JCorrect
   • C
     $69$ J

     Why not C: Used $\cos 30°$ instead of $\sin 30°$ for height.
   • D
     $80$ J

     Why not D: Used full $d=4$ m as vertical height.
   Explanation

   Height fallen: $h=d\sin 30°=4\times 0.5=2$ m. Work by gravity $=mgh=2\times 10\times 2=40$ J. No friction, so $\Delta KE=40$ J. Starting from rest, $K{E}_f=40$ J.

   Key takeaway

   On a frictionless incline, $W_{gravity} = mgh = mgd\sin\theta$. This equals the gain in kinetic energy.

5. Question 5 · Medium

   A variable force $F(x)=3x^2$ N acts on a particle along the $x$-axis. What is the work done from $x=0$ to $x=2$ m?

   • A
     $6$ J

     Why not A: Evaluated $F(2)\cdot 2=12\cdot 0.5$, treating it as constant.
   • B
     $8$ JCorrect
   • C
     $12$ J

     Why not C: Computed $F(2)\times d=12\times 1$ without integrating.
   • D
     $4$ J

     Why not D: Used $F(1)\times 2=6\times \frac{2}{3}$ with a midpoint estimation error.
   Explanation

   $W={\int }_0^{2}F(x)dx={\int }_0^{2}3x^{2}dx={\left[x^3\right]}_0^2=8-0=8$ J.

   Key takeaway

   For a variable force, $W = \int_{x_i}^{x_f} F(x)\,dx$. Constant-force shortcuts don't apply.

6. Question 6 · Medium

   A 5 kg block is released from rest at the top of a frictionless ramp of height $h=3$ m and then slides along a rough horizontal surface (${\mu }_k=0.2$, $g=10$ m/s²). How far does it travel on the rough surface before stopping?

   • A
     $3$ m

     Why not A: Set stopping distance equal to the ramp height.
   • B
     $7.5$ m

     Why not B: Used ${\mu }_k=0.4$ instead of $0.2$ when computing friction work.
   • C
     $15$ mCorrect
   • D
     $30$ m

     Why not D: Used $mgh/{\mu }_k$ without dividing by $mg$ (dimensional error).
   Explanation

   At the bottom of the ramp: $KE=mgh=5\times 10\times 3=150$ J. On the rough surface friction does work $W_f=-{\mu }_{k}mgd=-0.2\times 5\times 10\times d=-10d$. Setting $KE+W_f=0$: $150-10d=0\Rightarrow d=15$ m.

   Key takeaway

   Use energy conservation from the top of the ramp to find $KE$ at the bottom, then set $\mu_k mg \cdot d = KE$ to find the stopping distance on the rough surface.

7. Question 7 · Medium

   A motor lifts a 200 kg load at a constant velocity of $2$ m/s. What power does the motor deliver? ($g=10$ m/s²)

   • A
     $400$ W

     Why not A: Computed $mg/v$ instead of $mgv$.
   • B
     $1000$ W

     Why not B: Used $m\times v=400$ and multiplied by $g/2$.
   • C
     $4000$ WCorrect
   • D
     $2000$ W

     Why not D: Forgot $g$: computed $m\times v=400$ W.
   Explanation

   At constant velocity, $F_{lift}=mg=200\times 10=2000$ N. Power: $P=Fv=2000\times 2=4000$ W.

   Key takeaway

   $P = Fv$ when force and velocity are parallel. At constant velocity, the lifting force equals weight.

8. Question 8 · Medium

   A spring ($k=500$ N/m) launches a 0.2 kg ball from rest after being compressed 0.2 m. Using energy conservation, find the ball's launch speed.

   • A
     $5$ m/s

     Why not A: Used $v=kx/m$ (dimensional mismatch) instead of the energy equation.
   • B
     $\sqrt{50}\approx 7.07$ m/s

     Why not B: Forgot the $\frac{1}{2}$ on the spring-energy side: used $k{x}^2/m$ instead of $k{x}^2/(2m)$.
   • C
     $\sqrt{100}=10$ m/sCorrect
   • D
     $\sqrt{200}\approx 14.1$ m/s

     Why not D: Omitted the $\frac{1}{2}$ on both sides and then double-counted.
   Explanation

   Energy conservation: $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2\Rightarrow v=x\sqrt{\frac{k}{m}}=0.2\sqrt{\frac{500}{0.2}}=0.2\sqrt{2500}=0.2\times 50=10$ m/s.

   Key takeaway

   Spring-to-kinetic energy conversion: $v = x\sqrt{k/m}$. The $\frac{1}{2}$ cancels from both sides, leaving $v = x\sqrt{k/m}$.

9. Question 9 · Medium

   A 4 kg block moves along the $x$-axis subject to force $F(x)=6x-2$ N. Starting from rest at $x=1$ m, find the kinetic energy at $x=3$ m.

   • A
     $16$ J

     Why not A: Computed ${\int }_1^3(6x)dx$ only, omitting the $-2$ term.
   • B
     $20$ JCorrect
   • C
     $24$ J

     Why not C: Evaluated $F(3)\times 2=16\times 2=32$; over-estimated.
   • D
     $12$ J

     Why not D: Used average $F$ at endpoints without integrating: $\bar{F}=\frac{4+16}{2}=10$, $W=10\times 1.2$.
   Explanation

   $W={\int }_1^3(6x-2)dx={\left[3x^2-2x\right]}_1^3=(27-6)-(3-2)=21-1=20$ J. By the work–energy theorem: $K{E}_f=0+20=20$ J.

   Key takeaway

   With a variable force, compute work as $\int F(x)\,dx$, then apply $W_{net} = \Delta KE$.

10. Question 10 · Hard

    A conservative force on a particle is $F(x)=-\frac{dU}{dx}$ where $U(x)=2x^3-x^2$ J. Find the force at $x=1$ m.

    • A
      $-4$ NCorrect
    • B
      $4$ N

      Why not B: Forgot the negative sign in $F=-dU/dx$.
    • C
      $-2$ N

      Why not C: Differentiated $2x^3$ as $2x^2$ and $-x^2$ as $-2x$ without the full chain rule coefficient.
    • D
      $1$ J (wrong unit)

      Why not D: Evaluated $U(1)=1$ and treated potential energy as force.
    Explanation

    $\frac{dU}{dx}=6x^2-2x$. At $x=1$: $dU/dx=6-2=4$ J/m $=$ N. Force: $F=-\frac{dU}{dx}=-4$ N.

    Key takeaway

    For a conservative force, $F = -dU/dx$. Differentiate the potential energy function and negate.

11. Question 11 · Hard

    Using the work integral, derive the work done by a spring force $F_s=-kx$ as a block is displaced from $x=0$ to $x=A$.

    • A
      $W=-kA$

      Why not A: Used $W=F\cdot d=(-kA)(A)$ without integrating (treated force as constant).
    • B
      $W=-\frac{1}{2}k{A}^2$Correct
    • C
      $W=\frac{1}{2}k{A}^2$

      Why not C: Correct magnitude but wrong sign: spring force opposes displacement.
    • D
      $W=k{A}^2$

      Why not D: Forgot the $\frac{1}{2}$ factor from integration and the negative sign.
    Explanation

    $W_{spring}={\int }_0^A{F}_{s}dx={\int }_0^A(-kx)dx={\left[-\frac{k}{2}{x}^2\right]}_0^A=-\frac{1}{2}k{A}^2$.

    The spring does negative work on the block as it moves away from equilibrium (the stored energy $U_s=+\frac{1}{2}k{A}^2$ must come from the external agent, not the spring itself).

    Key takeaway

    The spring does work $W_s = -\frac{1}{2}kA^2$ (negative) when stretched/compressed by $A$. The equal and opposite elastic PE stored is $U_s = +\frac{1}{2}kA^2$.

12. Question 12 · Hard

    A particle moves in the $xy$-plane from $(0,0)$ to $(2,4)$ m under the force $\mathbf{F}=(3x^2)\hat{i}+(2y)\hat{j}$ N along the straight-line path $y=2x$. What is the total work done?

    • A
      $8$ J

      Why not A: Computed only $\int F_{x}dx$ from $0$ to $2$.
    • B
      $16$ J

      Why not B: Computed only $\int F_{y}dy$ from $0$ to $4$.
    • C
      $24$ JCorrect
    • D
      $32$ J

      Why not D: Added wrong intermediate results or doubled one integral.
    Explanation

    $W=\int \mathbf{F}\cdot d\mathbf{r}={\int }_0^{2}3x^{2}dx+{\int }_0^{4}2ydy$.

    ${\int }_0^{2}3x^{2}dx={\left[x^3\right]}_0^2=8$ J. ${\int }_0^{4}2ydy={\left[y^2\right]}_0^4=16$ J. Total $W=8+16=24$ J.

    Note: Since the curl $\nabla \times \mathbf{F}=0$ ($\partial F_x/\partial y=0$ and $\partial F_y/\partial x=0$), $\mathbf{F}$ is conservative and the result is path-independent.

    Key takeaway

    For 2D work, $W = \int F_x\,dx + \int F_y\,dy$, separating into $x$- and $y$-components. If $\mathbf{F}$ is conservative, the path doesn't matter.