Translational Kinetic Energy - AP Physics C Mechanics

1. Identify the unit of kinetic energy in the SI system.

   Joule (J). Energy unit in metric system, equivalent to newton-meter.

2. Calculate kinetic energy: $m = 8 \text{ kg}$, $v = 7 \text{ m/s}$.

   $KE = 196 \text{ J}$. Using $KE = \frac{1}{2}(8)(7^2) = \frac{1}{2}(8)(49) = 196$ J.

3. Express kinetic energy in terms of momentum $p$ and mass $m$.

   $KE = \frac{p^2}{2m}$. Derived from $p = mv$ and substituting into kinetic energy formula.

4. What is the kinetic energy of a $2 \text{ kg}$ object at $5 \text{ m/s}$?

   $KE = 25 \text{ J}$. Using $KE = \frac{1}{2}(2)(5^2) = \frac{1}{2}(2)(25) = 25$ J.

5. What variable represents velocity in the kinetic energy formula?

   $v$. Velocity is the vector quantity representing speed and direction.

6. Calculate the mass: $KE = 18 \text{ J}$, $v = 3 \text{ m/s}$.

   $m = 4 \text{ kg}$. Solving $18 = \frac{1}{2}m(3^2)$ gives $m = 4$ kg.

7. Calculate the velocity: $KE = 36 \text{ J}$, $m = 4 \text{ kg}$.

   $v = 4.24 \text{ m/s}$. Solving $36 = \frac{1}{2}(4)v^2$ gives $v = \sqrt{18} ≈ 4.24$ m/s.

8. How does kinetic energy change if mass is halved?

   Halves. KE is directly proportional to mass.

9. What is the kinetic energy of a $3 \text{ kg}$ object at $0 \text{ m/s}$?

   $KE = 0 \text{ J}$. Zero velocity means object has no kinetic energy.

10. What is the dimensional formula for kinetic energy?

    $ML^2T^{-2}$. Mass times length squared per time squared.

11. If an object's kinetic energy is zero, what can be said about its velocity?

    Velocity is zero. Zero kinetic energy requires zero velocity (assuming non-zero mass).

12. Express kinetic energy in terms of work done.

    Work done = Change in kinetic energy. Work-energy theorem relates work done to kinetic energy change.

13. Determine the velocity: $KE = 50 \text{ J}$, $m = 5 \text{ kg}$.

    $v = 4.47 \text{ m/s}$. Solving $50 = \frac{1}{2}(5)v^2$ gives $v = \sqrt{20} ≈ 4.47$ m/s.

14. What is the kinetic energy change if velocity doubles?

    Increases by a factor of 4. KE varies with $v^2$, so doubling velocity quadruples energy.

15. Express the kinetic energy formula using $F$, $d$, and $m$.

    No direct formula; use $KE = \frac{1}{2}mv^2$. Must use velocity relationship; force and displacement alone insufficient.

16. What is the formula for translational kinetic energy?

    $KE = \frac{1}{2}mv^2$. Standard formula where energy equals half mass times velocity squared.

17. What happens to kinetic energy if velocity is constant?

    KE remains constant. Constant velocity means no change in kinetic energy.

18. What is the relationship between kinetic energy and speed?

    KE is proportional to $v^2$. Kinetic energy has quadratic dependence on velocity.

19. Calculate $p$: $KE = 20 \text{ J}$, $m = 4 \text{ kg}$.

    $p = 8.94 \text{ kg m/s}$. Using $p = \sqrt{2mKE} = \sqrt{2(4)(20)} = \sqrt{160} ≈ 8.94$ kg⋅m/s.

20. If an object's velocity triples, how does its kinetic energy change?

    Increases by a factor of 9. KE varies with $v^2$, so tripling velocity increases energy ninefold.

21. Determine $v$: $KE = 45 \text{ J}$, $m = 3 \text{ kg}$.

    $v = 5.48 \text{ m/s}$. Solving $45 = \frac{1}{2}(3)v^2$ gives $v = \sqrt{30} ≈ 5.48$ m/s.

22. What is the relationship between kinetic energy and work done?

    Work done equals change in kinetic energy. Work-energy theorem connects work done to kinetic energy change.

23. Determine the kinetic energy of a $10 \text{ kg}$ object moving at $5 \text{ m/s}$.

    $KE = 125 \text{ J}$. Using $KE = \frac{1}{2}(10)(5^2) = \frac{1}{2}(10)(25) = 125$ J.

24. Calculate kinetic energy: $m = 6 \text{ kg}$, $v = 3 \text{ m/s}$.

    $KE = 27 \text{ J}$. Using $KE = \frac{1}{2}(6)(3^2) = \frac{1}{2}(6)(9) = 27$ J.

25. What is the kinetic energy of a $1 \text{ kg}$ object at rest?

    $KE = 0 \text{ J}$. Object at rest has zero velocity, therefore zero kinetic energy.

26. What is the kinetic energy when $m = 0 \text{ kg}$?

    $KE = 0 \text{ J}$. Zero mass means no kinetic energy regardless of velocity.

27. If kinetic energy is constant, what happens to mass if velocity increases?

    Mass decreases. If KE constant and $v$ increases, $m$ must decrease proportionally.

28. Calculate the mass: $KE = 100 \text{ J}$, $v = 10 \text{ m/s}$.

    $m = 2 \text{ kg}$. Solving $100 = \frac{1}{2}m(10^2)$ gives $m = 2$ kg.

29. What is the kinetic energy when $v = 0 \text{ m/s}$?

    $KE = 0 \text{ J}$. Zero velocity means object is at rest, no kinetic energy.

30. Calculate the kinetic energy: $m = 2 \text{ kg}$, $v = 3 \text{ m/s}$.

    $KE = 9 \text{ J}$. Using $KE = \frac{1}{2}(2)(3^2) = \frac{1}{2}(2)(9) = 9$ J.

31. What is the kinetic energy change for a velocity change from $2 \text{ m/s}$ to $4 \text{ m/s}$?

    Increases by a factor of 4. Velocity doubles from 2 to 4, so KE increases fourfold.

32. Differentiate kinetic energy with respect to velocity.

    $\frac{d(KE)}{dv} = mv$. Derivative of $\frac{1}{2}mv^2$ with respect to velocity gives momentum.

33. How does kinetic energy change if velocity doubles?

    Increases by a factor of 4. KE varies with $v^2$, so doubling velocity quadruples energy.

34. What variable represents mass in the kinetic energy formula?

    $m$. Mass is the scalar quantity representing object's inertia.

35. What is the kinetic energy of a $5 \text{ kg}$ object at $10 \text{ m/s}$?

    $KE = 250 \text{ J}$. Using $KE = \frac{1}{2}(5)(10^2) = \frac{1}{2}(5)(100) = 250$ J.

36. What is the effect on kinetic energy if both mass and velocity double?

    Increases by a factor of 8. Mass doubles KE, velocity squared quadruples it: $2 × 4 = 8$.

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