Graph Energy and Speed - Middle School Physical Science

1. Which point must lie on any $KE$ versus $v$ graph for an object (mass constant)?

   $(0,0)$. When speed is zero, kinetic energy must be zero.

2. If speed doubles from $v$ to $2v$, by what factor does $KE$ change (mass constant)?

   It becomes $4$ times as large. Since $KE \propto v^2$: $(2v)^2 = 4v^2$.

3. State the formula for kinetic energy in terms of mass $m$ and speed $v$.

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

4. Calculate $KE$ when $m = 1\,\text{kg}$ and $v = 4\,\text{m/s}$.

   $8\,\text{J}$. $KE = \frac{1}{2}(1)(4^2) = \frac{1}{2}(1)(16) = 8\,\text{J}$

5. Choose the correct statement: $KE$ increases linearly with $v$ or quadratically with $v$?

   Quadratically with $v$. The $v^2$ term makes the relationship quadratic.

6. Which proportionality correctly describes kinetic energy versus speed for constant mass?

   $KE \propto v^2$. From $KE = \frac{1}{2}mv^2$, kinetic energy varies with speed squared.

7. Compute $KE$ for $m = 4\,\text{kg}$ and $v = 2\,\text{m/s}$.

   $8\,\text{J}$. $KE = \frac{1}{2}(4)(2)^2 = \frac{1}{2}(4)(4) = 8\,\text{J}$.

8. On a graph of $KE$ versus $v^2$, what is the slope in terms of mass $m$?

   Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m(v^2)$, slope equals the coefficient.

9. Calculate $KE$ when $m = 2\,\text{kg}$ and $v = 3\,\text{m/s}$.

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

10. Choose the point that must lie on a $KE$ vs $v$ graph for constant $m$: $(0,0)$ or $(0,1)$.

    $(0,0)$. At rest ($v=0$), kinetic energy must be zero.

11. Identify which plot is linear: $KE$ vs $v$ or $KE$ vs $v^2$ (for constant mass).

    $KE$ vs $v^2$. Since $KE \propto v^2$, plotting against $v^2$ gives a straight line.

12. Identify which graph is correct for constant mass: linear $KE$ vs $v$ or curved $KE$ vs $v$.

    Curved $KE$ vs $v$. The $v^2$ relationship creates a curve, not a straight line.

13. Which option best describes the relationship between $KE$ and $v$ for constant mass?

    $KE \propto v^2$. This notation means $KE$ is proportional to velocity squared.

14. Identify the correct axes for a graph showing how kinetic energy changes with speed.

    $x$-axis: $v$; $y$-axis: $KE$. Speed is the independent variable, kinetic energy is dependent.

15. What is the general shape of a $KE$ versus $v$ graph for constant mass?

    An upward-opening curve (parabola). Since $KE \propto v^2$, the graph forms a parabola.

16. State the formula for kinetic energy $KE$ in terms of mass $m$ and speed $v$.

    $KE = \frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.

17. What happens to the slope of a $KE$ versus $v$ graph as $v$ increases (mass constant)?

    It increases (graph gets steeper). The parabola's slope increases as you move right.

18. For $m = 1\,\text{kg}$, list $KE$ values for $v = 0,1,2,3\,\text{m/s}$.

    $0,\frac{1}{2},2,\frac{9}{2}\,\text{J}$. Apply $KE = \frac{1}{2}(1)v^2$ for each speed value.

19. If you graph $KE$ versus $v^2$ (mass constant), what shape should the graph be?

    A straight line. $KE = \frac{1}{2}m(v^2)$ shows direct proportionality.

20. Identify the best graph type to show how $KE$ changes as speed changes continuously.

    A line graph (smooth curve). Continuous data requires a smooth curve, not discrete points.

21. Identify the correct graph transformation: if mass doubles, how does the $KE$ vs $v$ graph change?

    All $KE$ values double (vertical stretch by $2$). Doubling $m$ doubles $KE$ at every speed since $KE \propto m$.

22. Identify how kinetic energy changes when speed doubles and mass stays constant.

    It becomes $4$ times as large. Since $KE \propto v^2$, doubling $v$ gives $2^2 = 4$ times the energy.

23. Identify the missing point: if $m$ is constant and $KE = 2\,\text{J}$ at $v = 1$, what is $KE$ at $v = 3$?

    $18\,\text{J}$. Since $KE \propto v^2$, at $v=3$: $KE = 2 \times 3^2 = 18\,\text{J}$

24. Identify how kinetic energy changes when speed triples and mass stays constant.

    It becomes $9$ times as large. Since $KE \propto v^2$, tripling $v$ gives $3^2 = 9$ times the energy.

25. Identify the dependent variable when graphing kinetic energy versus speed.

    Kinetic energy $KE$ (y-axis). The variable that responds to changes goes on the y-axis.

26. Identify the independent variable when graphing kinetic energy versus speed.

    Speed $v$ (x-axis). The variable you control/change goes on the x-axis.

27. Identify the missing point: if $m$ is constant and $KE = 50\,\text{J}$ at $v = 5$, what is $KE$ at $v = 10$?

    $200\,\text{J}$. Since $KE \propto v^2$, doubling $v$ quadruples $KE$: $50 \times 4 = 200\,\text{J}$

28. Identify how kinetic energy changes when speed is cut in half and mass stays constant.

    It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$, halving $v$ gives $(\frac{1}{2})^2 = \frac{1}{4}$ times the energy.

29. What is the shape of a $KE$ versus speed $v$ graph for a constant mass object?

    An upward-opening curve (parabola). The $v^2$ relationship creates a parabolic curve.

30. Calculate $KE$ when $m = 4\,\text{kg}$ and $v = 2\,\text{m/s}$.

    $8\,\text{J}$. $KE = \frac{1}{2}(4)(2^2) = \frac{1}{2}(4)(4) = 8\,\text{J}$

31. If speed is cut in half from $v$ to $\frac{1}{2}v$, by what factor does $KE$ change?

    It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$: $(\frac{1}{2}v)^2 = \frac{1}{4}v^2$.

32. Find the correct statement: if $v$ increases from $2$ to $6$ with constant $m$, by what factor does $KE$ change?

    It increases by a factor of $9$. Speed ratio is $\frac{6}{2} = 3$, so $KE$ ratio is $3^2 = 9$.

33. If speed triples from $v$ to $3v$, by what factor does $KE$ change (mass constant)?

    It becomes $9$ times as large. Since $KE \propto v^2$: $(3v)^2 = 9v^2$.

34. Compute $KE$ for $m = 2\,\text{kg}$ and $v = 3\,\text{m/s}$.

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

35. Which axis label is correct for a standard $KE$ versus $v$ graph in SI units?

    x: $v\,(\text{m/s})$, y: $KE\,(\text{J})$. Speed in m/s on x-axis, energy in joules on y-axis.

36. For constant speed, how does kinetic energy change if mass doubles from $m$ to $2m$?

    It doubles. $KE$ is directly proportional to mass when speed is constant.

37. Identify the correct graph description for constant mass: straight line or upward-curving line for $KE$ vs $v$?

    Upward-curving line for $KE$ vs $v$. Quadratic relationships produce curved (parabolic) graphs.

38. State the value of kinetic energy when an object has speed $v = 0$.

    $KE = 0$. No motion means no kinetic energy.

39. Identify the graph feature that shows $KE$ increases faster as $v$ increases.

    The curve gets steeper as $v$ increases. The parabola's increasing slope shows accelerating growth.

Study these flashcards