Energy Speed Patterns - Middle School Physical Science
Card 1 of 25
What happens to kinetic energy when speed triples and mass stays the same?
What happens to kinetic energy when speed triples and mass stays the same?
Tap to reveal answer
It becomes $9$ times larger. Since $KE \propto v^2$, tripling $v$ gives $(3v)^2 = 9v^2$.
It becomes $9$ times larger. Since $KE \propto v^2$, tripling $v$ gives $(3v)^2 = 9v^2$.
← Didn't Know|Knew It →
Choose the correct unit for kinetic energy on a graph axis: $\text{J}$ or $\text{N}$.
Choose the correct unit for kinetic energy on a graph axis: $\text{J}$ or $\text{N}$.
Tap to reveal answer
$\text{J}$. Joules (J) is the SI unit for energy; Newtons (N) measure force.
$\text{J}$. Joules (J) is the SI unit for energy; Newtons (N) measure force.
← Didn't Know|Knew It →
A $KE$ vs. $v$ graph is a straight line through the origin. Identify the most likely error.
A $KE$ vs. $v$ graph is a straight line through the origin. Identify the most likely error.
Tap to reveal answer
It should be $KE$ vs. $v^2$, not $v$. A straight line suggests linear relationship, but $KE$ varies with $v^2$, not $v$.
It should be $KE$ vs. $v^2$, not $v$. A straight line suggests linear relationship, but $KE$ varies with $v^2$, not $v$.
← Didn't Know|Knew It →
A graph shows $KE$ quadruples when $v$ doubles. What relationship does this indicate?
A graph shows $KE$ quadruples when $v$ doubles. What relationship does this indicate?
Tap to reveal answer
$KE \propto v^2$. Quadrupling when doubling confirms the squared relationship.
$KE \propto v^2$. Quadrupling when doubling confirms the squared relationship.
← Didn't Know|Knew It →
Identify the correct pattern: if $v$ increases by $1\ \text{m/s}$ each step, does $KE$ increase by a constant amount?
Identify the correct pattern: if $v$ increases by $1\ \text{m/s}$ each step, does $KE$ increase by a constant amount?
Tap to reveal answer
No; the increases get larger each step. Since $KE \propto v^2$, equal increases in $v$ produce increasingly larger $KE$ changes.
No; the increases get larger each step. Since $KE \propto v^2$, equal increases in $v$ produce increasingly larger $KE$ changes.
← Didn't Know|Knew It →
Find $v$ if $m = 2\ \text{kg}$ and $KE = 16\ \text{J}$.
Find $v$ if $m = 2\ \text{kg}$ and $KE = 16\ \text{J}$.
Tap to reveal answer
$v = 4\ \text{m/s}$. From $16 = \frac{1}{2}(2)v^2$, solving gives $v^2 = 16$, so $v = 4$ m/s.
$v = 4\ \text{m/s}$. From $16 = \frac{1}{2}(2)v^2$, solving gives $v^2 = 16$, so $v = 4$ m/s.
← Didn't Know|Knew It →
Find $KE$ for $m = 2\ \text{kg}$ and $v = 3\ \text{m/s}$.
Find $KE$ for $m = 2\ \text{kg}$ and $v = 3\ \text{m/s}$.
Tap to reveal answer
$KE = 9\ \text{J}$. Using $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9$ J.
$KE = 9\ \text{J}$. Using $KE = \frac{1}{2}(2)(3)^2 = \frac{1}{2}(2)(9) = 9$ J.
← Didn't Know|Knew It →
Two objects have the same kinetic energy. Which has greater speed: the one with larger $m$ or smaller $m$?
Two objects have the same kinetic energy. Which has greater speed: the one with larger $m$ or smaller $m$?
Tap to reveal answer
The one with smaller $m$. From $v = \sqrt{\frac{2KE}{m}}$, smaller $m$ gives larger $v$ when $KE$ is constant.
The one with smaller $m$. From $v = \sqrt{\frac{2KE}{m}}$, smaller $m$ gives larger $v$ when $KE$ is constant.
← Didn't Know|Knew It →
Two objects have the same speed. Which has greater $KE$: the one with larger $m$ or smaller $m$?
Two objects have the same speed. Which has greater $KE$: the one with larger $m$ or smaller $m$?
Tap to reveal answer
The one with larger $m$. From $KE = \frac{1}{2}mv^2$, larger $m$ gives larger $KE$ when $v$ is constant.
The one with larger $m$. From $KE = \frac{1}{2}mv^2$, larger $m$ gives larger $KE$ when $v$ is constant.
← Didn't Know|Knew It →
Which object has greater mass if both lines are on a $KE$ vs. $v^2$ graph?
Which object has greater mass if both lines are on a $KE$ vs. $v^2$ graph?
Tap to reveal answer
The object with the steeper slope. Steeper slope means larger $\frac{1}{2}m$, therefore larger mass.
The object with the steeper slope. Steeper slope means larger $\frac{1}{2}m$, therefore larger mass.
← Didn't Know|Knew It →
On a $KE$ vs. $v$ graph, which point must be on the curve for any object?
On a $KE$ vs. $v$ graph, which point must be on the curve for any object?
Tap to reveal answer
$(0,0)$. When $v = 0$, $KE = \frac{1}{2}m(0)^2 = 0$, so the curve passes through the origin.
$(0,0)$. When $v = 0$, $KE = \frac{1}{2}m(0)^2 = 0$, so the curve passes through the origin.
← Didn't Know|Knew It →
A $KE$ vs. $v^2$ line has slope $3$. What is the mass $m$ in kilograms?
A $KE$ vs. $v^2$ line has slope $3$. What is the mass $m$ in kilograms?
Tap to reveal answer
$m = 6\ \text{kg}$. Since slope $= \frac{1}{2}m = 3$, solving gives $m = 6$ kg.
$m = 6\ \text{kg}$. Since slope $= \frac{1}{2}m = 3$, solving gives $m = 6$ kg.
← Didn't Know|Knew It →
Identify the slope of a $KE$ vs. $v^2$ graph for an object of mass $m$.
Identify the slope of a $KE$ vs. $v^2$ graph for an object of mass $m$.
Tap to reveal answer
Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.
Slope $= \frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is $\frac{1}{2}m$.
← Didn't Know|Knew It →
Which graph becomes a straight line for constant mass: $KE$ vs. $v$ or $KE$ vs. $v^2$?
Which graph becomes a straight line for constant mass: $KE$ vs. $v$ or $KE$ vs. $v^2$?
Tap to reveal answer
$KE$ vs. $v^2$. Since $KE = \frac{1}{2}m \cdot v^2$, plotting against $v^2$ gives a linear relationship.
$KE$ vs. $v^2$. Since $KE = \frac{1}{2}m \cdot v^2$, plotting against $v^2$ gives a linear relationship.
← Didn't Know|Knew It →
Which graph shape best represents $KE$ vs. $v$ for a constant mass: line or curve?
Which graph shape best represents $KE$ vs. $v$ for a constant mass: line or curve?
Tap to reveal answer
Curve (upward-opening parabola). The $v^2$ relationship creates a parabolic curve, not a straight line.
Curve (upward-opening parabola). The $v^2$ relationship creates a parabolic curve, not a straight line.
← Didn't Know|Knew It →
Identify the relationship between $KE$ and $v$ when mass is constant.
Identify the relationship between $KE$ and $v$ when mass is constant.
Tap to reveal answer
$KE \propto v^2$. Since $m$ is constant in $KE = \frac{1}{2}mv^2$, $KE$ is proportional to $v^2$.
$KE \propto v^2$. Since $m$ is constant in $KE = \frac{1}{2}mv^2$, $KE$ is proportional to $v^2$.
← Didn't Know|Knew It →
State the formula for kinetic energy in terms of mass $m$ and speed $v$.
State the formula for kinetic energy in terms of mass $m$ and speed $v$.
Tap to reveal answer
$KE = \frac{1}{2}mv^2$. Shows kinetic energy equals half the product of mass and velocity squared.
$KE = \frac{1}{2}mv^2$. Shows kinetic energy equals half the product of mass and velocity squared.
← Didn't Know|Knew It →
If mass doubles and speed stays constant, by what factor does kinetic energy change?
If mass doubles and speed stays constant, by what factor does kinetic energy change?
Tap to reveal answer
$2\times$. KE is directly proportional to mass when speed is constant.
$2\times$. KE is directly proportional to mass when speed is constant.
← Didn't Know|Knew It →
Which graph shape best represents kinetic energy vs speed for constant mass: linear, quadratic, or inverse?
Which graph shape best represents kinetic energy vs speed for constant mass: linear, quadratic, or inverse?
Tap to reveal answer
Quadratic (upward-curving parabola). The $v^2$ relationship creates a parabolic curve opening upward.
Quadratic (upward-curving parabola). The $v^2$ relationship creates a parabolic curve opening upward.
← Didn't Know|Knew It →
What is the formula for kinetic energy of a moving object in terms of mass and speed?
What is the formula for kinetic energy of a moving object in terms of mass and speed?
Tap to reveal answer
$KE=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.
$KE=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.
← Didn't Know|Knew It →
On a $KE$ vs $v$ graph for constant mass, what is the kinetic energy when $v=0$?
On a $KE$ vs $v$ graph for constant mass, what is the kinetic energy when $v=0$?
Tap to reveal answer
$0\ \text{J}$. When velocity is zero, there's no motion and no kinetic energy.
$0\ \text{J}$. When velocity is zero, there's no motion and no kinetic energy.
← Didn't Know|Knew It →
On a $KE$ vs $v^2$ graph, what does the slope equal for an object of mass $m$?
On a $KE$ vs $v^2$ graph, what does the slope equal for an object of mass $m$?
Tap to reveal answer
$\frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is the slope.
$\frac{1}{2}m$. From $KE = \frac{1}{2}m \cdot v^2$, the coefficient of $v^2$ is the slope.
← Didn't Know|Knew It →
Which graph becomes linear if you plot kinetic energy on the y-axis and $v^2$ on the x-axis (mass constant)?
Which graph becomes linear if you plot kinetic energy on the y-axis and $v^2$ on the x-axis (mass constant)?
Tap to reveal answer
$KE$ vs $v^2$. Since $KE = \frac{1}{2}m \cdot v^2$, plotting vs $v^2$ gives a straight line.
$KE$ vs $v^2$. Since $KE = \frac{1}{2}m \cdot v^2$, plotting vs $v^2$ gives a straight line.
← Didn't Know|Knew It →
Identify the pattern: as speed increases, do equal increases in speed add equal increases in kinetic energy?
Identify the pattern: as speed increases, do equal increases in speed add equal increases in kinetic energy?
Tap to reveal answer
No; kinetic energy increases by larger amounts. Due to the $v^2$ term, each speed increase adds more KE than the last.
No; kinetic energy increases by larger amounts. Due to the $v^2$ term, each speed increase adds more KE than the last.
← Didn't Know|Knew It →
If kinetic energy increases by a factor of $16$ with constant mass, by what factor did speed change?
If kinetic energy increases by a factor of $16$ with constant mass, by what factor did speed change?
Tap to reveal answer
$4\times$. If $KE$ increases $16\times$, then $v$ increases $\sqrt{16} = 4\times$.
$4\times$. If $KE$ increases $16\times$, then $v$ increases $\sqrt{16} = 4\times$.
← Didn't Know|Knew It →