Graph Energy and Mass - Middle School Physical Science

1. If $KE = 40\,J$ at $m = 5\,kg$, what is the constant speed $v$?

   $v = 4\,m/s$. From $40 = \frac{1}{2}(5)v^2$, we get $v^2 = 16$, so $v = 4\,m/s$.

2. Calculate $KE$ for $m=4\,\text{kg}$ and $v=3\,\text{m/s}$.

   $18\,\text{J}$. $KE=\frac{1}{2}(4)(3)^2=\frac{1}{2}(4)(9)=18\,\text{J}$

3. If $v$ is constant at $4\,\text{m/s}$, what is the slope of the $KE$ vs. $m$ graph?

   $8\,\text{J/kg}$. Slope $= \frac{1}{2}v^2 = \frac{1}{2}(4)^2 = \frac{1}{2}(16) = 8\,\text{J/kg}$.

4. At constant $v = 6\,m/s$, what is $KE$ when $m = 0.5\,kg$?

   $9\,J$. $KE = \frac{1}{2}(0.5)(6^2) = \frac{1}{2}(0.5)(36) = 9\,J$.

5. Find the slope from two points: $(m,KE)=(1,5)$ and $(3,15)$ on a $KE$ vs. $m$ graph.

   $5\,\text{J/kg}$. Slope $=\frac{\Delta KE}{\Delta m}=\frac{15-5}{3-1}=\frac{10}{2}=5$ J/kg.

6. If $v$ is constant and a point on the graph is $(m,KE)=(2,10)$, what is $KE$ at $m=6$?

   $30$. Since $KE \propto m$, if $m$ triples from 2 to 6, $KE$ triples from 10 to 30.

7. Which option correctly labels a $KE$ vs. $m$ graph for constant $v$ in SI units?

   $x$: $m$ (kg), $y$: $KE$ (J). Proper axis labeling includes variable symbols and units.

8. What variable should be on the $x$-axis to graph how $KE$ changes with mass at constant speed?

   Mass, $m$. Independent variable goes on horizontal axis when graphing relationships.

9. Which variable should be on the $x$-axis to graph how $KE$ changes with mass at constant $v$?

   Mass $m$ on the $x$-axis. The independent variable (what we change) goes on the horizontal axis.

10. Choose the correct axis labels for a $KE$ vs $m$ graph at constant $v$.

    x-axis: $m$ (kg); y-axis: $KE$ (J). Labels show variable name and units in parentheses.

11. What is the $y$-intercept of a $KE$ versus $m$ graph when $v$ is constant and $m=0$ is included?

    $0$. When mass is zero, kinetic energy must also be zero.

12. What is the kinetic energy formula you use before graphing $KE$ versus mass?

    $KE=\frac{1}{2}mv^2$. This formula relates kinetic energy to mass and velocity squared.

13. What is the $KE$-axis intercept on a $KE$ vs. $m$ graph when $v$ is constant?

    $0$. When $m = 0$, $KE = \frac{1}{2}(0)v^2 = 0$, so the line passes through $(0,0)$.

14. Identify the correct ordered pair format for a $KE$ vs. $m$ graph at constant $v$.

    Plot points as $(m,\,KE)$. Standard notation places the independent variable first, dependent variable second.

15. At constant $v = 4\,m/s$, what is $KE$ when $m = 3\,kg$?

    $24\,J$. $KE = \frac{1}{2}(3)(4^2) = \frac{1}{2}(3)(16) = 24\,J$.

16. If $v$ is constant and the slope of a $KE$ vs. $m$ graph is $8$, what is $v$?

    $4\,\text{m/s}$. Since slope $= \frac{1}{2}v^2 = 8$, then $v^2 = 16$, so $v = 4\,\text{m/s}$.

17. 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 product of mass and velocity squared.

18. What is the graph shape of $KE$ versus $m$ when speed $v$ is constant?

    A straight line through the origin. Direct proportionality always produces a linear graph through $(0,0)$.

19. Find the missing point: if $v$ is constant and $KE = 10\,\text{J}$ at $m = 2\,\text{kg}$, what is $KE$ at $m = 5\,\text{kg}$?

    $25\,\text{J}$. Since $KE \propto m$, if mass increases by factor of $2.5$, so does $KE$: $10 \times 2.5 = 25\,\text{J}$.

20. Calculate $KE$ for $m = 2\,kg$ and $v = 3\,m/s$.

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

21. Which variable should be on the $y$-axis to graph how $KE$ changes with mass at constant $v$?

    Kinetic energy $KE$ on the $y$-axis. The dependent variable (what we measure) goes on the vertical axis.

22. Identify the correct label for the $y$-axis including units when graphing $KE$ vs. $m$.

    $KE\,(\text{J})$. The $y$-axis shows the dependent variable with its units.

23. State the slope of a $KE$ versus $m$ graph when speed $v$ is constant.

    Slope $=\frac{1}{2}v^2$. From $KE=\frac{1}{2}mv^2$, the coefficient of $m$ is the slope.

24. What is the formula for kinetic energy in terms of mass $m$ and speed $v$?

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

25. State the slope of a $KE$ vs. $m$ graph when speed $v$ is constant.

    Slope $=\frac{1}{2}v^2$. From $KE = \frac{1}{2}mv^2$, when graphing $KE$ vs. $m$, the coefficient of $m$ is the slope.

26. What is the kinetic energy formula for an object of mass $m$ moving at speed $v$?

    $KE = \frac{1}{2}mv^2$. Energy equals half the product of mass and velocity squared.

27. If mass doubles while speed $v$ stays constant, how does $KE$ change?

    $KE$ doubles. Direct proportionality means if $m$ doubles, $KE$ doubles too.

28. If $v$ is constant, what is $KE$ at $m=0\,\text{kg}$ on a $KE$ vs. $m$ graph?

    $0\,\text{J}$. No mass means no kinetic energy, regardless of speed.

29. What unit should be used on the $y$-axis when graphing kinetic energy $KE$?

    Joules ($\text{J}$). The SI unit for energy is the joule, defined as $\text{kg} \cdot \text{m}^2/\text{s}^2$.

30. Find the missing mass: at constant $v=3\,\text{m/s}$, a point has $KE=9\,\text{J}$. What is $m$?

    $2\,\text{kg}$. $9 = \frac{1}{2}m(9)$, so $m = 2$ kg.

31. Identify the speed from a $KE$ vs $m$ slope of $8\,\text{J/kg}$ using $\text{slope}=\frac{1}{2}v^2$.

    $4\,\text{m/s}$. Since slope $=\frac{1}{2}v^2=8$, then $v^2=16$, so $v=4\,\text{m/s}$.

32. What happens to $KE$ if mass doubles while speed stays constant?

    $KE$ doubles. Direct proportionality means $KE$ scales linearly with mass.

33. Choose the correct equation form for a $KE$ vs. $m$ line at constant speed $v$.

    $KE = \left(\frac{1}{2}v^2\right)m$. This form shows $KE$ as a linear function of $m$ with slope $\frac{1}{2}v^2$.

34. Which variable is the dependent variable when graphing how $KE$ changes with object mass?

    Kinetic energy, $KE$. KE changes in response to mass, making it dependent.

35. If mass triples while speed stays constant, by what factor does $KE$ change?

    It triples (factor of $3$). Since $KE \propto m$, tripling mass triples kinetic energy.

36. Identify the correct ordered pair format for a $KE$ vs $m$ graph at constant $v$.

    $(m,\,KE)$. Ordered pairs show (x-coordinate, y-coordinate).

37. If $v$ is constant, what is the $y$-intercept of a $KE$ vs. $m$ graph (ideal model)?

    $0$ (passes through the origin). When $m = 0$, $KE = 0$, so the line passes through the origin.

38. Calculate $KE$ for $m=4\,\text{kg}$ and $v=3\,\text{m/s}$ to plot a point on a graph.

    $18\,\text{J}$. $KE=\frac{1}{2}(4)(3^2)=\frac{1}{2}(4)(9)=18$ J.

39. If $v$ is constant and a point on the graph is $(m,KE)=(8,40)$, what is $KE$ at $m=2$?

    $10$. Since $KE \propto m$, if $m$ is divided by 4 (from 8 to 2), $KE$ is divided by 4 (40 to 10).

40. Choose the correct statement about the graph: at constant $v$, increasing $m$ makes $KE$ increase at a constant rate.

    True; the $KE$ vs $m$ graph is linear. Linear relationship means constant rate of change.

41. Which variable is the independent variable when graphing how $KE$ changes with object mass?

    Mass, $m$. Mass is what we're changing to see its effect on KE.

42. Identify the correct unit for kinetic energy to label the $y$-axis on a graph.

    Joules, $J$. SI unit for energy, where $1\,J = 1\,kg \cdot m^2/s^2$.

43. If speed is constant, what is the $KE$ ratio when mass changes from $m$ to $3m$?

    $KE$ triples (ratio $3:1$). If mass triples and $v$ is constant, $KE$ also triples due to direct proportionality.

44. Identify the graph shape of $KE$ versus mass $m$ when speed $v$ is constant.

    A straight line through the origin (directly proportional). Since $KE = \frac{1}{2}mv^2$ and $v$ is constant, $KE$ is directly proportional to $m$.

45. Calculate $KE$ for $m = 4\,kg$ and $v = 2\,m/s$.

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

46. Choose the correct point for a $KE$ vs. $m$ graph at $v = 3\,m/s$ when $m = 2\,kg$.

    $(2,\,9)$. When $m = 2\,kg$ and $v = 3\,m/s$, $KE = \frac{1}{2}(2)(9) = 9\,J$.

47. Which axis should mass $m$ be placed on when graphing $KE$ as a function of mass?

    The $x$-axis. Independent variables go on the horizontal axis.

48. What happens to $KE$ when mass $m$ increases while speed $v$ stays constant?

    $KE$ increases in direct proportion to $m$. Since $KE = \frac{1}{2}mv^2$ and $v$ is constant, $KE$ is directly proportional to $m$.

49. Which variable should be on the $x$-axis when graphing how $KE$ changes with mass (constant $v$)?

    Mass $m$ on the $x$-axis. Independent variable (mass) goes on the horizontal axis.

50. What is the relationship between $KE$ and mass $m$ when speed $v$ is constant?

    $KE\propto m$ (directly proportional). From $KE=\frac{1}{2}mv^2$, KE varies linearly with mass when $v$ is fixed.

51. Choose the correct units for the slope of a $KE$ vs. $m$ graph: $\text{J/kg}$ or $\text{kg/J}$.

    $\text{J/kg}$. Slope units are $\frac{\text{J}}{\text{kg}}$ from $\frac{\Delta KE}{\Delta m}$.

52. Which variable belongs on the $x$-axis when graphing how $KE$ changes with mass at constant $v$?

    Mass $m$. The independent variable (what we're changing) goes on the horizontal axis.

53. If $v$ is constant, what is the slope of a $KE$ vs. $m$ graph in terms of $v$?

    Slope $= \frac{1}{2}v^2$. From $KE = \frac{1}{2}v^2 \cdot m$, the coefficient of $m$ is the slope.

54. What is $KE$ for $m = 4\,\text{kg}$ and $v = 3\,\text{m/s}$?

    $18\,\text{J}$. Using $KE = \frac{1}{2}(4)(3)^2 = \frac{1}{2}(4)(9) = 18\,\text{J}$.

55. If $v$ is constant, find $KE$ when $m=5\,\text{kg}$ and slope is $4\,\text{J/kg}$.

    $20\,\text{J}$. $KE=$ slope $\times$ mass $= 4 \times 5 = 20$ J.

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

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

57. What happens to $KE$ if mass is cut in half while speed stays constant?

    $KE$ is cut in half. KE is directly proportional to mass, so halving $m$ halves KE.

58. What happens to $KE$ if mass triples while speed stays constant?

    $KE$ triples. Linear relationship means $KE$ increases by same factor as mass.

59. If a $KE$ vs. $m$ graph has slope $18$, what constant speed $v$ does it imply?

    $v = 6\,m/s$. Since slope $= \frac{1}{2}v^2 = 18$, then $v^2 = 36$, so $v = 6\,m/s$.

60. Identify the correct unit for mass to label the $x$-axis on a graph.

    Kilograms, $kg$. SI base unit for mass in the metric system.

61. Which graph is correct for constant $v$: line through origin, horizontal line, or curve upward?

    Line through the origin. Direct proportionality means a straight line passing through the origin.

62. At constant $v=6\,\text{m/s}$, what is $KE$ for $m=2\,\text{kg}$?

    $36\,\text{J}$. $KE = \frac{1}{2}(2)(6^2) = \frac{1}{2}(2)(36) = 36$ J.

63. Identify the y-intercept of a $KE$ vs $m$ graph when speed $v$ is constant.

    $0\,\text{J}$. When $m=0$, $KE=\frac{1}{2}(0)v^2=0$.

64. If $v$ is constant, what is the expected shape of a graph of $KE$ (y) versus $m$ (x)?

    A straight line through the origin. Direct proportionality produces a linear graph through (0,0).

65. At constant $v=4\,\text{m/s}$, what is $KE$ for $m=3\,\text{kg}$?

    $24\,\text{J}$. $KE = \frac{1}{2}(3)(4^2) = \frac{1}{2}(3)(16) = 24$ J.

66. Calculate $KE$ for $m=4\,\text{kg}$ and $v=3\,\text{m/s}$.

    $18\,\text{J}$. $KE = \frac{1}{2}(4)(3^2) = \frac{1}{2}(4)(9) = 18\,\text{J}$

67. Identify the relationship between $KE$ and $m$ when speed $v$ is constant.

    $KE\propto m$ (directly proportional). Since $KE=\frac{1}{2}mv^2$, when $v$ is constant, $KE$ increases linearly with $m$.

68. Identify the correct label for the $x$-axis including units when graphing $KE$ vs. $m$.

    $m\,(\text{kg})$. The $x$-axis shows the independent variable with its units.

69. Identify the correct SI units for mass $m$ used on a $KE$ vs. $m$ graph.

    Kilograms (kg). Mass is measured in kilograms in the SI system.

70. Calculate $KE$ for $m=2\,\text{kg}$ and $v=3\,\text{m/s}$ to plot a point on a graph.

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

71. At constant $v = 5\,m/s$, what is $KE$ when $m = 2\,kg$?

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

72. If a $KE$ vs. $m$ graph has slope $8$, what constant speed $v$ does it imply?

    $v = 4\,m/s$. Since slope $= \frac{1}{2}v^2 = 8$, then $v^2 = 16$, so $v = 4\,m/s$.

73. What is the slope of a $KE$ vs. $m$ graph when speed $v$ is constant?

    Slope $=\frac{1}{2}v^2$. From $KE=\frac{1}{2}mv^2$, the coefficient of $m$ is the slope.

74. If $v$ is constant, what is the proportional relationship between $KE$ and mass $m$?

    $KE \propto m$. Direct proportionality since $KE = \frac{1}{2}mv^2$ with constant $v$.

75. Which equation matches a $KE$ vs $m$ graph at constant $v$ written as $KE=km$?

    $k=\frac{1}{2}v^2$. Comparing $KE=km$ with $KE=\frac{1}{2}mv^2$ shows $k=\frac{1}{2}v^2$.

76. What is the independent variable when graphing how $KE$ changes with mass $m$ at constant $v$?

    Mass $m$ (x-axis). The variable we control/change goes on the x-axis.

77. What is the $y$-intercept of a $KE$ vs. $m$ graph if $v$ is constant and $KE = \frac{1}{2}mv^2$?

    $0$. When $m = 0$, $KE = \frac{1}{2}(0)v^2 = 0$, so the line passes through the origin.

78. Identify the correct proportionality for constant speed: is $KE \propto m$, $KE \propto m^2$, or $KE \propto \frac{1}{m}$?

    $KE \propto m$. From $KE = \frac{1}{2}mv^2$, at constant $v$, kinetic energy varies linearly with mass.

79. What is the expected graph shape of $KE$ vs. $m$ when $v$ is constant?

    A straight line through the origin. Direct proportionality produces a linear graph passing through $(0,0)$.

80. Find the missing value: at constant $v$, $m=5\,\text{kg}$ gives $KE=20\,\text{J}$; what is $KE$ at $m=10\,\text{kg}$?

    $40\,\text{J}$. Since $KE\propto m$, doubling mass doubles kinetic energy.

81. What is the expected graph shape of $KE$ vs. $m$ when speed $v$ is constant?

    A straight line (linear increase). Since $KE = \frac{1}{2}mv^2$ and $v$ is constant, $KE$ increases linearly with $m$.

82. Identify the correct units for kinetic energy $KE$ used on a graph in SI units.

    Joules (J). Energy is measured in joules in the SI system.

83. Which variable belongs on the $y$-axis when graphing how $KE$ changes with mass at constant $v$?

    Kinetic energy $KE$. The dependent variable (what we're measuring) goes on the vertical axis.

84. What does the slope represent on a $KE$ (y) versus $m$ (x) graph when $v$ is constant?

    $\frac{1}{2}v^2$. From $KE = \frac{1}{2}mv^2$, slope equals the constant $\frac{1}{2}v^2$.

85. What happens to $KE$ if mass $m$ doubles while speed $v$ stays the same?

    $KE$ doubles. Since $KE \propto m$ when $v$ is constant, doubling $m$ doubles $KE$.

86. What unit should be used on the $x$-axis when graphing mass $m$?

    Kilograms ($\text{kg}$). The SI unit for mass is the kilogram.

87. A line on a $KE$ vs $m$ graph has slope $8\,\text{J/kg}$. What is the speed $v$?

    $4\,\text{m/s}$. Slope = $\frac{1}{2}v^2 = 8$, so $v^2 = 16$ and $v = 4$ m/s.

88. What is the dependent variable when graphing how $KE$ changes with mass $m$ at constant $v$?

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

89. Identify the mistake: A student graphs $KE$ vs. $m$ but changes $v$ for each mass value.

    Speed $v$ must be held constant. Changing $v$ violates the controlled variable requirement for this graph.

90. What is $KE$ for $m = 2\,\text{kg}$ and $v = 3\,\text{m/s}$?

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

91. If $v=4\,\text{m/s}$, what is the slope of the $KE$ vs. $m$ graph?

    $8\,\text{J/kg}$. Slope $=\frac{1}{2}v^2=\frac{1}{2}(4^2)=\frac{1}{2}(16)=8$ J/kg.

92. Identify the correct ordered-pair format for plotting a $KE$ versus $m$ graph at constant $v$.

    $(m,\,KE)$. Mass is x-coordinate, KE is y-coordinate in ordered pairs.

93. What variable should be on the $y$-axis to graph how kinetic energy changes with mass?

    Kinetic energy, $KE$. Dependent variable goes on vertical axis to show how it changes.

94. Calculate $KE$ for $m=4\,\text{kg}$ and $v=3\,\text{m/s}$.

    $18\,\text{J}$. $KE = \frac{1}{2}(4)(3^2) = \frac{1}{2}(4)(9) = 18$ J.

95. 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 product of mass and velocity squared.

96. If the slope of a $KE$ vs. $m$ graph is $18\,\text{J/kg}$, what is the speed $v$?

    $6\,\text{m/s}$. From slope $=\frac{1}{2}v^2$: $18=\frac{1}{2}v^2$, so $v^2=36$, thus $v=6$ m/s.

97. Calculate $KE$ for $m=2\,\text{kg}$ and $v=5\,\text{m/s}$.

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

98. Compute the slope $\frac{\Delta KE}{\Delta m}$ from points $(2,10)$ and $(6,30)$ on a $KE$ vs $m$ graph.

    $5\,\text{J/kg}$. Slope $=\frac{30-10}{6-2}=\frac{20}{4}=5\,\text{J/kg}$

99. Which variable should be on the $y$-axis when graphing how $KE$ changes with mass (constant $v$)?

    Kinetic energy $KE$ on the $y$-axis. Dependent variable (KE) goes on the vertical axis.

100. Choose the correct conclusion if a $KE$ vs. $m$ graph is a line through the origin (constant $v$).

     $KE$ increases linearly with $m$. A line through origin indicates direct proportionality.

101. If speed is constant, what is the $KE$ ratio when mass changes from $m$ to $2m$?

     $KE$ doubles (ratio $2:1$). If mass doubles and $v$ is constant, $KE$ also doubles due to direct proportionality.

102. Which axis should kinetic energy $KE$ be placed on when graphing $KE$ as a function of mass?

     The $y$-axis. Dependent variables go on the vertical axis.

103. If mass triples while speed $v$ stays constant, how does $KE$ change?

     $KE$ triples. Direct proportionality means if $m$ triples, $KE$ triples too.

104. What is the relationship between $KE$ and $m$ when speed $v$ is held constant?

     $KE\propto m$. Direct proportionality means KE increases linearly with mass.

105. What does the slope of a $KE$ vs. $m$ graph equal when speed $v$ is constant?

     Slope $= \frac{1}{2}v^2$. From $KE = \frac{1}{2}mv^2$, when graphed as $KE$ vs. $m$, the coefficient of $m$ is the slope.

106. Which set of points matches $v=2\,\text{m/s}$ for $m=1,2,3\,\text{kg}$ on a $KE$ vs $m$ graph?

     $(1,2),(2,4),(3,6)$ (with $KE$ in J). At $v=2$ m/s: $KE = \frac{1}{2}m(4)$, giving 2, 4, 6 J for m = 1, 2, 3 kg.

107. Find the missing value: at constant $v$, $m=8\,\text{kg}$ gives $KE=32\,\text{J}$; what is $KE$ at $m=2\,\text{kg}$?

     $8\,\text{J}$. Since $KE\propto m$, one-fourth the mass gives one-fourth the energy.

108. Which of these points must lie on a correct $KE$ vs. $m$ graph at constant $v$: $(0,0)$ or $(0,5)$?

     $(0,0)$. A direct proportion line must pass through the origin $(0,0)$.

109. What happens to $KE$ if mass $m$ is cut in half while speed $v$ stays the same?

     $KE$ is cut in half. Since $KE \propto m$ when $v$ is constant, halving $m$ halves $KE$.

110. Identify the correct plot points for $v=2\,\text{m/s}$ at $m=0,1,2\,\text{kg}$ on a $KE$ vs $m$ graph.

     $(0,0)$, $(1,2)$, $(2,4)$. Using $KE=\frac{1}{2}mv^2$ with $v=2$: $(0,0)$, $(1,2)$, $(2,4)$.

111. Calculate $KE$ for $m=2\,\text{kg}$ and $v=5\,\text{m/s}$.

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

112. If mass doubles while speed stays constant, by what factor does $KE$ change?

     It doubles (factor of $2$). Since $KE \propto m$, doubling mass doubles kinetic energy.

113. Which graph is correct for constant $v$: a line through origin or a curve that bends upward?

     A line through the origin. Linear relationships with direct proportionality always pass through the origin.

114. Identify the correct proportional relationship between $KE$ and $m$ when $v$ is constant.

     $KE\propto m$. From $KE = \frac{1}{2}mv^2$ with constant $v$, $KE$ is directly proportional to $m$.

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