Apply Work-Energy Theorem - Physics

1. What is the work done by the normal force on an object moving along a surface it is perpendicular to?

   $W_N=0$. Normal force is perpendicular to motion, so $\cos(90°) = 0$.

2. What is the spring potential energy for displacement $x$ from equilibrium?

   $U_s=\frac{1}{2}kx^2$. Elastic potential energy is quadratic in displacement.

3. What is the energy form associated with gravity near Earth, using height $y$?

   $U_g=mgy$. Gravitational potential energy increases with height.

4. Identify the formula for final speed when $W_{\text{net}}$ is known: start at $v_i$, mass $m$.

   $v_f=\sqrt{v_i^2+\frac{2W_{\text{net}}}{m}}$. Derived from $W_{\text{net}} = \frac{1}{2}m(v_f^2 - v_i^2)$.

5. What is the formula for translational kinetic energy of a mass $m$ moving at speed $v$?

   $K=\frac{1}{2}mv^2$. Kinetic energy depends on mass and the square of velocity.

6. What is the SI unit of work and energy?

   $1\ \text{J}=1\ \text{N}\cdot\text{m}$. Joule equals newton-meter, the unit of work and energy.

7. Find $W_g$ for $m=5\ \text{kg}$ rising by $\Delta y=2\ \text{m}$ with $g=9.8\ \text{m/s}^2$.

   $W_g=-98\ \text{J}$. $W_g = -mg\Delta y = -(5)(9.8)(2) = -98$ J for upward motion.

8. Find $W_{\text{net}}$ if $m=2\ \text{kg}$, $v_i=3\ \text{m/s}$, and $v_f=7\ \text{m/s}$.

   $W_{\text{net}}=40\ \text{J}$. $\Delta K = \frac{1}{2}(2)(49-9) = 40$ J using work-energy theorem.

9. What is the work done by the gravitational force when an object changes height by $\Delta y$?

   $W_g=-mg\Delta y$. Negative because gravity opposes upward displacement.

10. Find the work by gravity for $m=3\ \text{kg}$ rising by $\Delta y=2\ \text{m}$ with $g=9.8\ \text{m/s}^2$.

    $-58.8\ \text{J}$. $W_g = -3 \times 9.8 \times 2 = -58.8$ J.

11. Find the work by kinetic friction if $f_k=6\ \text{N}$ acts over $d=4\ \text{m}$.

    $W_f=-24\ \text{J}$. Friction opposes motion: $W_f = -(6)(4) = -24$ J.

12. What is the work done by a constant force $F$ over displacement $d$ at angle $\theta$?

    $W=Fd\cos\theta$. $\theta$ is the angle between force and displacement vectors.

13. Find the work done by a $10\ \text{N}$ force over $3\ \text{m}$ when $\theta=60^\circ$.

    $W=15\ \text{J}$. $W = 10 \times 3 \times \cos(60°) = 30 \times 0.5 = 15$ J.

14. What is the work done by a spring force when stretched from $x_1$ to $x_2$?

    $W_s = \frac{1}{2}k(x_1^2-x_2^2)$. Spring work depends on initial and final position squares.

15. What is the Work-Energy Theorem written using net work and kinetic energy?

    $W_{\text{net}}=\Delta K$. States that net work equals the change in kinetic energy.

16. What is the Work-Energy Theorem written as an equation for net work and kinetic energy?

    $W_{net} = \Delta K$. Net work equals the change in kinetic energy.

17. What is the normal force work on an object moving along a level surface with no vertical displacement?

    $W_N = 0$. Normal force perpendicular to motion does no work.

18. What is the SI unit of work and kinetic energy?

    $1\ \text{J} = 1\ \text{N}\cdot\text{m}$. Joule equals newton-meter, the unit of energy.

19. Find $W$ when $F=8\ \text{N}$, $d=5\ \text{m}$, and $\theta=60^\circ$.

    $20\ \text{J}$. $W = 8 \times 5 \times \cos(60°) = 40 \times 0.5 = 20$ J.

20. Find $v_f$ if $m=4\ \text{kg}$, $v_i=2\ \text{m/s}$, and $W_{net}=48\ \text{J}$.

    $5\ \text{m/s}$. $48 = \frac{1}{2}(4)(v_f^2-4)$; solving gives $v_f = 5$ m/s.

21. What is the work done by kinetic friction of magnitude $f_k$ over distance $d$?

    $W_f=-f_k d$. Friction always opposes motion, making work negative.

22. What is the work done by kinetic friction of magnitude $f_k$ over distance $d$?

    $W_f = -f_k d$. Friction always opposes motion, doing negative work.

23. Which sign does work have when the force component is opposite the displacement direction?

    $W<0$. Work is negative when force opposes motion ($\theta > 90°$).

24. Find $v_f$ if $m=4\ \text{kg}$, $v_i=0$, and $W_{\text{net}}=18\ \text{J}$.

    $v_f=3\ \text{m/s}$. $18 = \frac{1}{2}(4)v_f^2$ gives $v_f^2 = 9$, so $v_f = 3$ m/s.

25. Find the work by friction if $f_k=6\ \text{N}$ acts over $d=4\ \text{m}$ in the direction opposite motion.

    $-24\ \text{J}$. Friction opposes motion: $W_f = -6 \times 4 = -24$ J.

26. Find $W_{net}$ for $m=2\ \text{kg}$ when speed increases from $3\ \text{m/s}$ to $7\ \text{m/s}$.

    $40\ \text{J}$. $W_{net} = \frac{1}{2}(2)(49-9) = 40$ J.

27. What is the gravitational work done on mass $m$ for a vertical change $\Delta y$ (upward positive)?

    $W_g = -mg\Delta y$. Gravity does negative work when object rises.

28. Identify the work done by a force perpendicular to displacement (that is, $\theta = 90^\circ$).

    $W = 0$. Perpendicular force does no work since $\cos(90°)=0$.

29. State the formula for translational kinetic energy of a mass $m$ moving at speed $v$.

    $K = \frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.

30. What is the net work if speed changes from $v_i$ to $v_f$ for mass $m$?

    $W_{net} = \frac{1}{2}m(v_f^2-v_i^2)$. Apply work-energy theorem with initial and final speeds.

31. Find $W$ when $F=10\ \text{N}$, $d=3\ \text{m}$, and $\theta=0^\circ$.

    $30\ \text{J}$. $W = 10 \times 3 \times \cos(0°) = 30$ J.

32. Which option gives the sign of work when $\theta = 180^\circ$ between $\vec{F}$ and $\vec{d}$?

    $W<0$. Force opposite to displacement gives negative work.

33. Find $W_s$ for a spring with $k=200\ \text{N/m}$ moving from $x_i=0.30\ \text{m}$ to $x_f=0$.

    $W_s=9\ \text{J}$. $W_s = \frac{1}{2}(200)(0.09-0) = 9$ J as spring relaxes.

34. What is the definition of work done by a constant force $\vec{F}$ over displacement $\vec{d}$?

    $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$. Work is the dot product of force and displacement vectors.

35. Identify the net work if an object moves in a circle at constant speed (uniform circular motion).

    $W_{net}=0$. Constant speed means $\Delta K = 0$, so $W_{net} = 0$.

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