Graph Energy and Speed - Middle School Physical Science
Card 1 of 25
State the formula for kinetic energy $KE$ in terms of mass $m$ and speed $v$.
State the formula for kinetic energy $KE$ in terms of mass $m$ and speed $v$.
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$KE = \frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.
$KE = \frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.
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Identify the independent variable when graphing kinetic energy versus speed.
Identify the independent variable when graphing kinetic energy versus speed.
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Speed $v$ (x-axis). The variable you control/change goes on the x-axis.
Speed $v$ (x-axis). The variable you control/change goes on the x-axis.
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Identify the dependent variable when graphing kinetic energy versus speed.
Identify the dependent variable when graphing kinetic energy versus speed.
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Kinetic energy $KE$ (y-axis). The variable that responds to changes goes on the y-axis.
Kinetic energy $KE$ (y-axis). The variable that responds to changes goes on the y-axis.
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What is the general shape of a $KE$ versus $v$ graph for constant mass?
What is the general shape of a $KE$ versus $v$ graph for constant mass?
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An upward-opening curve (parabola). Since $KE \propto v^2$, the graph forms a parabola.
An upward-opening curve (parabola). Since $KE \propto v^2$, the graph forms a parabola.
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Which proportionality correctly describes kinetic energy versus speed for constant mass?
Which proportionality correctly describes kinetic energy versus speed for constant mass?
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$KE \propto v^2$. From $KE = \frac{1}{2}mv^2$, kinetic energy varies with speed squared.
$KE \propto v^2$. From $KE = \frac{1}{2}mv^2$, kinetic energy varies with speed squared.
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If speed doubles from $v$ to $2v$, by what factor does $KE$ change (mass constant)?
If speed doubles from $v$ to $2v$, by what factor does $KE$ change (mass constant)?
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It becomes $4$ times as large. Since $KE \propto v^2$: $(2v)^2 = 4v^2$.
It becomes $4$ times as large. Since $KE \propto v^2$: $(2v)^2 = 4v^2$.
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If speed triples from $v$ to $3v$, by what factor does $KE$ change (mass constant)?
If speed triples from $v$ to $3v$, by what factor does $KE$ change (mass constant)?
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It becomes $9$ times as large. Since $KE \propto v^2$: $(3v)^2 = 9v^2$.
It becomes $9$ times as large. Since $KE \propto v^2$: $(3v)^2 = 9v^2$.
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If speed is cut in half from $v$ to $\frac{1}{2}v$, by what factor does $KE$ change?
If speed is cut in half from $v$ to $\frac{1}{2}v$, by what factor does $KE$ change?
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It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$: $(\frac{1}{2}v)^2 = \frac{1}{4}v^2$.
It becomes $\frac{1}{4}$ as large. Since $KE \propto v^2$: $(\frac{1}{2}v)^2 = \frac{1}{4}v^2$.
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For constant speed, how does kinetic energy change if mass doubles from $m$ to $2m$?
For constant speed, how does kinetic energy change if mass doubles from $m$ to $2m$?
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It doubles. $KE$ is directly proportional to mass when speed is constant.
It doubles. $KE$ is directly proportional to mass when speed is constant.
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Compute $KE$ for $m = 2,\text{kg}$ and $v = 3,\text{m/s}$.
Compute $KE$ for $m = 2,\text{kg}$ and $v = 3,\text{m/s}$.
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$9,\text{J}$. $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9,\text{J}$.
$9,\text{J}$. $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9,\text{J}$.
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Compute $KE$ for $m = 4,\text{kg}$ and $v = 2,\text{m/s}$.
Compute $KE$ for $m = 4,\text{kg}$ and $v = 2,\text{m/s}$.
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$8,\text{J}$. $KE = \frac{1}{2}(4)(2)^2 = \frac{1}{2}(4)(4) = 8,\text{J}$.
$8,\text{J}$. $KE = \frac{1}{2}(4)(2)^2 = \frac{1}{2}(4)(4) = 8,\text{J}$.
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Which point must lie on any $KE$ versus $v$ graph for an object (mass constant)?
Which point must lie on any $KE$ versus $v$ graph for an object (mass constant)?
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$(0,0)$. When speed is zero, kinetic energy must be zero.
$(0,0)$. When speed is zero, kinetic energy must be zero.
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What happens to the slope of a $KE$ versus $v$ graph as $v$ increases (mass constant)?
What happens to the slope of a $KE$ versus $v$ graph as $v$ increases (mass constant)?
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It increases (graph gets steeper). The parabola's slope increases as you move right.
It increases (graph gets steeper). The parabola's slope increases as you move right.
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If you graph $KE$ versus $v^2$ (mass constant), what shape should the graph be?
If you graph $KE$ versus $v^2$ (mass constant), what shape should the graph be?
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A straight line. $KE = \frac{1}{2}m(v^2)$ shows direct proportionality.
A straight line. $KE = \frac{1}{2}m(v^2)$ shows direct proportionality.
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On a graph of $KE$ versus $v^2$, what is the slope in terms of mass $m$?
On a graph of $KE$ versus $v^2$, what is the slope in terms of mass $m$?
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Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m(v^2)$, slope equals the coefficient.
Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m(v^2)$, slope equals the coefficient.
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Which axis label is correct for a standard $KE$ versus $v$ graph in SI units?
Which axis label is correct for a standard $KE$ versus $v$ graph in SI units?
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x: $v,(\text{m/s})$, y: $KE,(\text{J})$. Speed in m/s on x-axis, energy in joules on y-axis.
x: $v,(\text{m/s})$, y: $KE,(\text{J})$. Speed in m/s on x-axis, energy in joules on y-axis.
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For $m = 1,\text{kg}$, list $KE$ values for $v = 0,1,2,3,\text{m/s}$.
For $m = 1,\text{kg}$, list $KE$ values for $v = 0,1,2,3,\text{m/s}$.
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$0,\frac{1}{2},2,\frac{9}{2},\text{J}$. Apply $KE = \frac{1}{2}(1)v^2$ for each speed value.
$0,\frac{1}{2},2,\frac{9}{2},\text{J}$. Apply $KE = \frac{1}{2}(1)v^2$ for each speed value.
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Choose the correct statement: $KE$ increases linearly with $v$ or quadratically with $v$?
Choose the correct statement: $KE$ increases linearly with $v$ or quadratically with $v$?
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Quadratically with $v$. The $v^2$ term makes the relationship quadratic.
Quadratically with $v$. The $v^2$ term makes the relationship quadratic.
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Identify the correct graph description for constant mass: straight line or upward-curving line for $KE$ vs $v$?
Identify the correct graph description for constant mass: straight line or upward-curving line for $KE$ vs $v$?
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Upward-curving line for $KE$ vs $v$. Quadratic relationships produce curved (parabolic) graphs.
Upward-curving line for $KE$ vs $v$. Quadratic relationships produce curved (parabolic) graphs.
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Identify the best graph type to show how $KE$ changes as speed changes continuously.
Identify the best graph type to show how $KE$ changes as speed changes continuously.
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A line graph (smooth curve). Continuous data requires a smooth curve, not discrete points.
A line graph (smooth curve). Continuous data requires a smooth curve, not discrete points.
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What is the shape of a $KE$ versus speed $v$ graph for a constant mass object?
What is the shape of a $KE$ versus speed $v$ graph for a constant mass object?
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An upward-opening curve (parabola). The $v^2$ relationship creates a parabolic curve.
An upward-opening curve (parabola). The $v^2$ relationship creates a parabolic curve.
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Identify the correct axes for a graph showing how kinetic energy changes with speed.
Identify the correct axes for a graph showing how kinetic energy changes with speed.
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$x$-axis: $v$; $y$-axis: $KE$. Speed is the independent variable, kinetic energy is dependent.
$x$-axis: $v$; $y$-axis: $KE$. Speed is the independent variable, kinetic energy is dependent.
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State the value of kinetic energy when an object has speed $v = 0$.
State the value of kinetic energy when an object has speed $v = 0$.
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$KE = 0$. No motion means no kinetic energy.
$KE = 0$. No motion means no kinetic energy.
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Identify the graph feature that shows $KE$ increases faster as $v$ increases.
Identify the graph feature that shows $KE$ increases faster as $v$ increases.
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The curve gets steeper as $v$ increases. The parabola's increasing slope shows accelerating growth.
The curve gets steeper as $v$ increases. The parabola's increasing slope shows accelerating growth.
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Which option best describes the relationship between $KE$ and $v$ for constant mass?
Which option best describes the relationship between $KE$ and $v$ for constant mass?
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$KE \propto v^2$. This notation means $KE$ is proportional to velocity squared.
$KE \propto v^2$. This notation means $KE$ is proportional to velocity squared.
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