Spring Forces - AP Physics C Mechanics

1. Determine the spring constant if $F_s = -10 \, \text{N}$ and $x = 0.25 \, \text{m}$.

   $k = 40 \, \text{N/m}$. From $F_s = -kx$, solve $k = -\frac{F_s}{x} = -\frac{-10}{0.25} = 40$ N/m.

2. Identify the phase of motion when a spring is at maximum extension.

   Maximum potential energy phase. All kinetic energy converted to potential energy at turning points.

3. What is the potential energy stored in a spring?

   $U = \frac{1}{2}kx^2$. Energy stored when spring is compressed or stretched from equilibrium.

4. Determine the potential energy if $x$ is zero.

   $U = 0 \, \text{J}$. At equilibrium position, no energy is stored in the spring.

5. Determine $k$ if $x = 0.4 \, \text{m}$ and $U = 6 \, \text{J}$.

   $k = 75 \, \text{N/m}$. From $U = \frac{1}{2}kx^2$, solve $k = \frac{2U}{x^2} = \frac{2(6)}{(0.4)^2} = 75$ N/m.

6. What happens to the spring constant if a spring is cut in half?

   Spring constant doubles. Shorter spring length increases stiffness proportionally.

7. Calculate $k$ if $F_s = -12 \, \text{N}$ and $x = -0.3 \, \text{m}$.

   $k = 40 \, \text{N/m}$. Calculate $k = -\frac{F_s}{x} = -\frac{-12}{-0.3} = 40$ N/m.

8. Find $F_s$ if $k = 120 \, \text{N/m}$ and $x = 0.05 \, \text{m}$.

   $F_s = -6 \, \text{N}$. Apply $F_s = -kx = -120 \times 0.05 = -6$ N.

9. What kind of motion results from a mass-spring system?

   Simple harmonic motion. Restoring force produces periodic oscillatory motion about equilibrium.

10. What is the work done by a spring force?

    Negative change in potential energy, $-\Delta U$. Conservative force work equals negative change in potential energy.

11. If $U = 2 \, \text{J}$ and $k = 80 \, \text{N/m}$, find $x$.

    $x = 0.2236 \, \text{m}$. Solve $x = \sqrt{\frac{2U}{k}} = \sqrt{\frac{2(2)}{80}} = 0.2236$ m.

12. Identify the unit of potential energy stored in a spring.

    Joules (J). Energy unit, same as work and all other forms of energy.

13. If $k = 50 \, \text{N/m}$ and $x = 0.3 \, \text{m}$, find $F_s$.

    $F_s = -15 \, \text{N}$. Calculate $F_s = -kx = -50 \times 0.3 = -15$ N.

14. What is the effect of mass on the period of a spring-mass system?

    Period increases with mass. Heavier mass results in slower oscillation with longer period.

15. What occurs when a spring is compressed?

    Potential energy increases. Compression stores elastic potential energy in the spring.

16. What is the relationship between spring potential energy and displacement?

    Quadratic relationship. Potential energy varies as the square of displacement.

17. If $k = 90 \, \text{N/m}$ and $x = 0.15 \, \text{m}$, find $U$.

    $U = 1.0125 \, \text{J}$. Calculate $U = \frac{1}{2}kx^2 = \frac{1}{2}(90)(0.15)^2 = 1.0125$ J.

18. State Hooke's Law formula for spring force.

    $F_s = -kx$. Force is proportional to displacement, with negative sign indicating restoring direction.

19. What is the effect of negative sign in Hooke's Law?

    Indicates force direction is opposite to displacement. Ensures spring always opposes displacement from equilibrium.

20. What is the relationship between spring force and acceleration?

    Acceleration is proportional to spring force. Newton's second law connects force directly to acceleration.

21. If $k = 75 \, \text{N/m}$ and $U = 4.5 \, \text{J}$, find $x$.

    $x = 0.3464 \, \text{m}$. Solve $x = \sqrt{\frac{2U}{k}} = \sqrt{\frac{2(4.5)}{75}} = 0.3464$ m.

22. Identify the type of energy conversion in a spring-mass system.

    Conversion between kinetic and potential energy. Energy alternates between kinetic and potential forms during oscillation.

23. What does the variable $x$ represent in Hooke's Law?

    Displacement from equilibrium position. Distance from the natural length where spring force equals zero.

24. What is the graphical representation of Hooke's Law?

    Linear graph of $F_s$ vs. $x$. Straight line with slope $-k$ passing through the origin.

25. What is the equilibrium position of a spring?

    Position where net force is zero. Natural length where spring exerts no force on attached objects.

26. What does the variable $k$ represent in Hooke's Law?

    Spring constant. Measures stiffness of the spring in the force-displacement relationship.

27. Determine $x$ if $U = 2.5 \, \text{J}$ and $k = 50 \, \text{N/m}$.

    $x = 0.3162 \, \text{m}$. Solve $x = \sqrt{\frac{2U}{k}} = \sqrt{\frac{2(2.5)}{50}} = 0.3162$ m.

28. What is the effect of increasing $k$ on the system's oscillation frequency?

    Oscillation frequency increases. Stiffer spring produces faster oscillations with higher frequency.

29. Determine the force if $k = 200 \, \text{N/m}$ and $x = 0.1 \, \text{m}$.

    $F_s = -20 \, \text{N}$. Apply $F_s = -kx = -200 \times 0.1 = -20$ N.

30. What does it mean if a spring is 'ideal'?

    Obeys Hooke's Law perfectly. Force is exactly proportional to displacement with no deviations.

31. What type of force is a spring force?

    Restorative force. Always acts to restore the spring to its equilibrium position.

32. How does the potential energy change if the spring constant doubles?

    Potential energy also doubles. Since $U = \frac{1}{2}kx^2$, doubling $k$ doubles the energy stored.

33. Identify the unit of the spring constant $k$.

    Newtons per meter (N/m). Force per unit displacement, combining force units (N) with length units (m).

34. Calculate the work done if $k=100 \, \text{N/m}$ and $x$ changes from $0.1 \, \text{m}$ to $0.2 \, \text{m}$.

    $W = -1.5 \, \text{J}$. $W = -\Delta U = -(\frac{1}{2}k(0.2)^2 - \frac{1}{2}k(0.1)^2) = -1.5$ J.

35. Calculate the potential energy for $k=150 \, \text{N/m}$ and $x=0.2 \, \text{m}$.

    $U = 3 \, \text{J}$. Use $U = \frac{1}{2}kx^2 = \frac{1}{2}(150)(0.2)^2 = 3$ J.

36. What happens to the spring force if displacement is halved?

    Spring force is also halved. Force is directly proportional to displacement per Hooke's Law.

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