A force $F_1 = ~f_1~ N$ is applied to move a box as shown in the figure. Find the work ($W_1$) done by $F_1$.
Hint: $W = \vec F \cdot \vec s= F s \cos \theta$.
Solution
Focus only on $\vec F_1$ and $\vec s$.
From the figure, the box moves to the right by $s = ~s~ m$.
Both $\vec s$ and $\vec F_1$ point to the right, so the angle between them is $\theta = 0^\circ$.
$$
\begin{eqnarray}
W_1 &=& \vec F_1 \cdot \vec s= F_1 s \cos \theta \\
&=& (~f_1~ N) (~s~ m) \cos [[return Math.abs(difference_in_angle_normalized(~theta_s~, ~theta_1~, false))]]^\circ \\
&=& [[return sf_latex(namespace_work.w_1)]] J
\end{eqnarray}
$$
Note that $1Nm = 1J$.
$W_1\gt 0$ means the force is giving energy to the box.
A box (or a train!) is sliding toward you. You apply a force $F_1 = ~f_1~ N$ to stop it as shown in the figure. Find the work ($W_1$) done by $F_1$.
Hint:
$W = \vec F \cdot \vec s= F s \cos \theta$.
Concerning Spidey's major: He is super nerdy.
Solution
Focus only on $\vec F_1$ and $\vec s$.
From the figure, the box moves to the left by $s = ~s~ m$.
$\vec s$ and $\vec F_1$ point in the opposite direction, so the angle between them is $\theta = 180^\circ$.
$$
\begin{eqnarray}
W_1 &=& \vec F_1 \cdot \vec s= F_1 s \cos \theta \\
&=& (~f_1~ N) (~s~ m) \cos [[return Math.abs(difference_in_angle_normalized(~theta_s~, ~theta_1~, false))]]^\circ \\
&=& [[return sf_latex(namespace_work.w_1)]] J
\end{eqnarray}
$$
Note that $1Nm = 1J$.
$W_1\lt 0$ means the force is taking energy from the box.
$W_1 = $
return sf_math(namespace_work.w_1);
5%
Select unit for work:
$J$
$m/s$
$N$
$m/s^2$
0
Does $F_1$ give or take energy away from the box?
Give
Take
1
What was Spider-Man's major in college?
Biology
Chemistry
Physics
Humanities
2
energy || work
Exercise - Positive or negative work in real life
Choose the sign of work as described below.
Solution
When $F$ gives energy, work is positive. $F$ is speeding up the object.
When $F$ takes energy away, work is negative. $F$ is slowing down the object.
Work done by you when you kick a ball.
Positive
Negative
Zero
0
Work done by the goal keeper when he stops a ball from scoring.
Positive
Negative
Zero
1
Work done by you as when you lift a heavy object.
Positive
Negative
Zero
0
Work done by gravity as when you lift a heavy object.
Positive
Negative
Zero
1
Work done by you as when you lower a heavy object.
Positive
Negative
Zero
1
Work done by air resistance as a raindrop falls.
Positive
Negative
Zero
1
Work done by gravity as a raindrop falls.
Positive
Negative
Zero
0
Work done by the normal force on a train as it heads downhill.
Positive
Negative
Zero
2
Work done by gravity on a train as it heads downhill.
Positive
Negative
Zero
0
Work done by friction on an ice skater.
Positive
Negative
Zero
1
work || energy
Exercise - Work done by $F$ in random directions
if (int_count_times_randomized == 0){
return 10.0;
} else {
return random_min_max_precision(5, 15, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 70;
} else {
return random_min_max_precision(-160, 160, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 14;
} else {
return random_min_max_precision(10, 15, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 30;
} else {
return random_min_max_precision(-160, 160, -1); //defined in setup_exercise_all.js
}
The block moves in the direction indicated as $s$.
A block of mass $m$ is released on a frictionless incline at angle $\theta$, tied to a rope with tension $T$ pulling on the block in the [[return (~tension~>0)? "uphill":"downhill"]] direction. The block is observed to move by distance $s$ [[return (~s~>0)? "uphill":"downhill"]] (note that $s$ does not have to be in the same direction as $T$).
Given:
$m = ~m~ kg$.
$T = [[return Math.abs(~tension~)]] N$.
$\theta = ~theta~ ^\circ$.
$s = [[return Math.abs(~s~)]] m$
Find:
The work done by normal force $W_n$.
The work done by gravity $W_g$.
The work done by the tension $W_T$.
The total work done on the block.
Hint:
Focus on the angle between a force and the displacement $s$.
$W_1 = F_1 s \cos \theta_1$: $\theta_1$ is the angle between $F$ and $s$, but is not necessarily the angle of the incline!
The force diagram.
Solution
Work done by the normal force
$F_n$ and $s$.
The angle between $F_n$ and $s$ is $90^\circ$:
$$
\begin{eqnarray}
W_n &=& F_n s \cos(90^\circ) \\
&=& 0 J
\end{eqnarray}
$$
Work done by gravity
The angle between $F_g$ and $s$ is $\theta_g = 90^\circ [[return (~s~>0)? "+":"-"]] ~theta~ ^\circ = [[return sf_latex(namespace_work.theta_g)]]^\circ$.
From the figure, we see that the angle between $F_g$ and $s$ is:
$$
\begin{eqnarray}
\theta_g &=& 90^\circ [[return (~s~>0)? "+":"-"]] \theta \\
&=& 90^\circ [[return (~s~>0)? "+":"-"]] ~theta~ ^\circ \\
&=& [[return sf_latex(namespace_work.theta_g)]]^\circ
\end{eqnarray}
$$
This gives:
$$
\begin{eqnarray}
W_g &=& (mg)(s)\cos\theta_g \\
&=& (~m~kg)(9.8 m/s^2)([[return Math.abs(~s~)]] m)\cos [[return sf_latex(namespace_work.theta_g)]]^\circ \\
&=& [[return sf_latex(namespace_work.w_g)]] J
\end{eqnarray}
$$
$W_g$ and $mgh$
Overall there are two cases:
Moving up: $W_g = -mgh$, so gravity takes energy from the block as it goes up, slowing it down.
Moving down: $W_g = +mgh$, so gravity gives the block energy as it slides down, speeding it up.
We are in the [[return (~s~>0)? "first":"second"]] case in this problem. You can randomize to see the other case.
where $h= s\sin\theta$ is the vertical height of the incline.
We can see this using the trig identity $\cos (90^\circ \pm \theta) = \mp \sin \theta$:
$$
\begin{eqnarray}
W_g &=& (mg)(s)\cos\theta_g \\
&=& (mg)(s)\cos (90^\circ [[return (~s~>0)? "+":"-"]] \theta) \\
&=& mg ([[return (~s~>0)? "-":"+"]] s \sin \theta) \\
&=& [[return (~s~>0)? "-":"+"]] mgh
\end{eqnarray}
$$
Click open "Alternative approach for $W_g$" below for a more physical explanation.
We could also break displacement into two components:
$$
\begin{eqnarray}
\vec s &=& s_x \hat i + s_y \hat j \\
&=& s \cos \theta \hat i [[return (~s~>0)? "+":"-"]] s \sin\theta \hat j \\
&=& ([[return sf_latex(~s~ * Math.cos(~theta~*Math.PI/180))]] \hat i [[return (~s~>0)? "+":""]] [[return sf_latex(~s~ * Math.sin(~theta~*Math.PI/180))]] \hat j) m
\end{eqnarray}
$$
Of the two components of $\vec s$, only the vertical component will contribute to $W_g$, so we can just ignore the $\hat i$ term.
There are two equivalent ways to proceed:
Method 1: Using $W = F s cos \theta$
The vertical component of $\vec s$ points [[return (~s~>0)? "up":"down"]], therefore it is [[return (~s~>0)?"anti-parallel":"paralle"]] to $F_g$, giving an angle of $[[return (~s~>0)? "180":"0"]]^\circ$ with $F_g$:
$$
\begin{eqnarray}
W_g &=& (mg)(|s_y|) \cos [[return (~s~>0)? "180":"0"]]^\circ \\
&=& (~m~kg)(9.8 m/s^2)([[return sf_latex(Math.abs(~s~ * Math.sin(~theta~*Math.PI/180)))]] m)\cos [[return (~s~>0)? "180":"0"]]^\circ \\
&=& [[return sf_latex(namespace_work.w_g)]] J
\end{eqnarray}
$$
If we don't put in numbers, we can write $|s_y| = h = s \sin \theta$:
$$
\begin{eqnarray}
W_g &=& (mg)(|s_y|) \cos [[return (~s~>0)? "180":"0"]]^\circ \\
&=& (mg)(h) ([[return (~s~>0)? "-":"+"]] 1) \\
&=& [[return (~s~>0)? "-":"+"]] mgh = [[return (~s~>0)? "-":"+"]] mg s \sin\theta
\end{eqnarray}
$$
where $\theta$ is the angle of the incline.
Method 2: Using $W = \vec F \cdot \vec s = F_x s_x + F_y s_y$
Weight is $\vec F_g = -mg \hat j = -[[return ~m~ * 9.8]] \hat j N$ in vector form (i.e. $F_x = 0$). Taking the dot product with $\vec s= ([[return sf_latex(~s~ * Math.cos(~theta~*Math.PI/180))]] \hat i [[return (~s~>0)? "+":""]] [[return sf_latex(~s~ * Math.sin(~theta~*Math.PI/180))]] \hat j) m$ gives:
$$
\begin{eqnarray}
W_g &=& 0 + (-[[return ~m~ * 9.8]] N)([[return sf_latex(~s~ * Math.sin(~theta~*Math.PI/180))]] m) \\
&=& [[return sf_latex(namespace_work.w_g)]] J
\end{eqnarray}
$$
Or symbolically, using $\vec F_g = -mg \hat j$ and $\vec s = s_x \hat i [[return (~s~>0)? "+":"-"]] h \hat j$ (where $h = |s_y|$), we get the dot product:
$$
\begin{eqnarray}
W_g &=& 0 + (-mg)([[return (~s~>0)? "+":"-"]] h) \\
&=& [[return (~s~>0)? "-":"+"]] mgh = [[return (~s~>0)? "-":"+"]] mg s \sin\theta
\end{eqnarray}
$$
Work done by tension
$T$ and $s$.
From the figure, we see that $T$ and $s$ are [[return (~s~*~tension~>0)?"parallel":"anti-parallel"]], so the angle between them is:
$$
\begin{eqnarray}
\theta_T &=& [[return sf_latex(namespace_work.theta_t)]]^\circ
\end{eqnarray}
$$
This gives:
$$
\begin{eqnarray}
W_T &=& (T)(s)\cos\theta_T \\
&=& ([[return Math.abs(~tension~)]] N)([[return Math.abs(~s~)]] m)\cos [[return sf_latex(namespace_work.theta_t)]]^\circ \\
&=& [[return sf_latex(namespace_work.w_t)]] J
\end{eqnarray}
$$
Total work
Sum up all three works we get:
$$
\begin{eqnarray}
W_{total} &=& W_n + W_g + W_T \\
&=& 0J + ([[return sf_latex(namespace_work.w_g)]] J) + ([[return sf_latex(namespace_work.w_t)]] J) \\
&=& [[return sf_latex(namespace_work.w_total)]] J
\end{eqnarray}
$$
A block of mass $m$ slides by distance $s$ on the horizontal ground with coefficient of friction $\mu$. A vertical force of magnitude $F_1$ is applied [[return (~theta_1~>0)? "upward":"downward"]] to the block as shown in the figure.
Given:
$m = ~m~ kg$.
$F_1 = |\vec F_1| = ~f_1~ N$.
$\mu = ~mu~$.
$s = ~s~ m$.
Draw the force diagram. (No online submission.)
Find the weight of the block, $F_g$.
Find the magnitude of the normal force, $F_n$.
Find the magnitude of friction, $f$.
Find the work done by friction $W_f$.
if (int_count_times_randomized == 0){
return 7.0;
} else {
return random_min_max_precision(6, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 40;
} else {
return random_min_max_precision(10, 50, 0, true); //defined in setup_exercise_all.js
}
We write the magnitudes of the upward and downward forces as $F_{up}$ and $F_{down}$ below.
Two upward forces: $F_n$ and $F_1$.
One downward force: $mg$.
No vertical motion (i.e. $a_y=0$) implies the $F_{up}$ must cancel $F_{down}$.
$$
\begin{eqnarray}
F_{up} &=& F_{down} \\
\Rightarrow F_n + F_1 &=& mg \\
\Rightarrow F_n &=& mg - F_1 \\
&=& (~m~ kg)(9.8 m/s^2) - ~f_1~ N \\
&=& [[return sf_latex(namespace_work.weight)]] N - ~f_1~ N \\
&=& [[return sf_latex(namespace_work.f_n)]] N
\end{eqnarray}
$$
$F_n \lt mg$ because $F_1$ helps support the weight of the block, so the ground does not need to push up as hard.
$F_1$ points down.
We write the magnitudes of the upward and downward forces as $F_{up}$ and $F_{down}$ below.
One upward force: $F_n$.
Two downward forces: $mg$ and $F_1$.
No vertical motion (i.e. $a_y=0$) implies the $F_{up}$ must cancel $F_{down}$.
$$
\begin{eqnarray}
F_{up} &=& F_{down} \\
\Rightarrow F_n &=& mg + F_1 \\
&=& (~m~ kg)(9.8 m/s^2) + ~f_1~ N \\
&=& [[return sf_latex(namespace_work.weight)]] N + ~f_1~ N \\
&=& [[return sf_latex(namespace_work.f_n)]] N
\end{eqnarray}
$$
$F_n \gt mg$ because $F_n$ must support both the weight of the block, and also cancel the downward press of $F_1$.
Friction
$$
\begin{eqnarray}
f &=& \mu F_n \\
&=& (~mu~)([[return sf_latex(namespace_work.f_n)]] N) \\
&=& [[return sf_latex(namespace_work.friction)]] N
\end{eqnarray}
$$
Work done by friction
The angle between $f$ and $s$ is $180^\circ$ because friction acts against the sliding motion of the block, giving:
$$
\begin{eqnarray}
W_f &=& f s \cos (180^\circ) \\
&=& ([[return sf_latex(namespace_work.friction)]] N) ( ~s~ m) (-1) \\
&=& [[return sf_latex(namespace_work.w_f)]] J
\end{eqnarray}
$$
$F_g = $
return sf_math(~m~ * 9.8);
5%
$F_n =$
return sf_math(Math.abs(namespace_work.f_n));
5%
$f =$
return sf_math(namespace_work.friction);
5%
$W_f =$
return sf_math(namespace_work.w_f);
5%
Select unit for force:
$kg$
$J$
$N$
$m/s^2$
2
Select unit for work:
$kg$
$J$
$N$
$m/s^2$
1
work || normal_force || friction
Exercise - Work on a rough horizontal surface with force at an angle
if (int_count_times_randomized == 0){
return 7.0;
} else {
return random_min_max_precision(6, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 40;
} else {
return random_min_max_precision(30, 50, 0, true); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 50;
} else {
return 2*random_min_max_precision(-40, 40, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 0.1;
} else {
return random_min_max_precision(0.1, 0.25, 2); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(3, 10, 0); //defined in setup_exercise_all.js
}
A block of mass $m$ moves by displacement $s$ horizontally on a rough surface with coefficient of kinetic friction $\mu$. A force of magnitude $F_1$ is applied at angle $\theta_1$ [[return (~theta_1~>=0)? "above":"below"]] the horizontal.
Find $W_{total}$, the total work done on the block.
Hint:
Find the vertical and horizontal components of $F_1$.
All the vertical forces must add up to zero because there is no vertical motion.
$f = \mu F_n$.
$W = F s \cos \theta$
Solution
The free body diagram.Decomposing $F_1$ (magnified).
$y$-components only.Friction $f$ and displacement $s$ are anti-parallel.
Work done by normal force and gravity
Both $F_n$ and $F_g$ are perpendicular to $s$, therefore:
$$
\begin{eqnarray}
W_n &=& F_n s \cos(90^\circ) = 0J \\
W_g &=& F_g s \cos(90^\circ) = 0J
\end{eqnarray}
$$
Work done by $F_1$
The angle between $F_1$ and $s$ is $\theta_1 = [[return Math.abs(~theta_1~)]]^\circ$, therefore:
$$
\begin{eqnarray}
W_1 &=& F_1 s \cos \theta_1 \\
&=& ([[return ~f_1~]] N)([[return Math.abs(~s~)]] m) \cos [[return Math.abs(~theta_1~)]]^\circ \\
&=& [[return sf_latex(namespace_work.w_1)]] J
\end{eqnarray}
$$
The normal force
$F_1$ has a vertical component $F_1\sin([[return Math.abs(~theta_1~)]]^\circ)$ pointing up.
We write the magnitudes of the upward and downward forces as $F_{up}$ and $F_{down}$ below.
Two upward forces: $F_n$ and $F_1\sin([[return Math.abs(~theta_1~)]]^\circ)$.
One downward force: $mg$.
No vertical motion (i.e. $a_y=0$) implies the $F_{up}$ must cancel $F_{down}$.
$$
\begin{eqnarray}
F_{up} &=& F_{down} \\
\Rightarrow F_n + F_1\sin([[return Math.abs(~theta_1~)]]^\circ) &=& mg \\
\Rightarrow F_n &=& mg - F_1\sin([[return Math.abs(~theta_1~)]]^\circ) \\
&=& (~m~ kg)(9.8 m/s^2) - ~f_1~ \sin([[return Math.abs(~theta_1~)]]^\circ) N \\
&=& [[return sf_latex(namespace_work.weight)]] N - [[return sf_latex(Math.abs(namespace_work.f_1_y))]] N \\
&=& [[return sf_latex(namespace_work.f_n)]] N
\end{eqnarray}
$$
$F_n \lt mg$ because $F_1$ helps support the weight of the block, so the ground does not need to push up as hard.
$F_1$ has a vertical component $F_1\sin([[return Math.abs(~theta_1~)]]^\circ)$ pointing down.
We write the magnitudes of the upward and downward forces as $F_{up}$ and $F_{down}$ below.
One upward force: $F_n$.
Two downward forces: $mg$ and $F_1\sin([[return Math.abs(~theta_1~)]]^\circ)$.
No vertical motion (i.e. $a_y=0$) implies the $F_{up}$ must cancel $F_{down}$.
$$
\begin{eqnarray}
F_{up} &=& F_{down} \\
\Rightarrow F_n &=& mg + F_1\sin([[return Math.abs(~theta_1~)]]^\circ) \\
&=& (~m~ kg)(9.8 m/s^2) + ~f_1~ \sin([[return Math.abs(~theta_1~)]]^\circ) N \\
&=& [[return sf_latex(namespace_work.weight)]] N + [[return sf_latex(Math.abs(namespace_work.f_1_y))]] N \\
&=& [[return sf_latex(namespace_work.f_n)]] N
\end{eqnarray}
$$
$F_n \gt mg$ because $F_n$ must support both the weight of the block, and also cancel the downward press of $F_1$.
$F_1$ has no vertical component.
We write the magnitudes of the upward and downward forces as $F_{up}$ and $F_{down}$ below.
One upward force: $F_n$.
One downward force: $mg$.
No vertical motion (i.e. $a_y=0$) implies the $F_{up}$ must cancel $F_{down}$.
$$
\begin{eqnarray}
F_{up} &=& F_{down} \\
\Rightarrow F_n &=& mg\\
&=& (~m~ kg)(9.8 m/s^2) \\
&=& [[return sf_latex(namespace_work.f_n)]] N
\end{eqnarray}
$$
$F_n = mg$ because $F_n$ supports the weight of the block and nothing else.
Friction
$$
\begin{eqnarray}
f &=& \mu F_n \\
&=& (~mu~ )([[return sf_latex(namespace_work.f_n)]] N) \\
&=& [[return sf_latex(namespace_work.friction)]] N
\end{eqnarray}
$$
Work done by friction
The angle between $f$ and $s$ is $180^\circ$ because friction acts against the sliding motion of the block, giving:
$$
\begin{eqnarray}
W_f &=& f s \cos (180^\circ) \\
&=& ([[return sf_latex(namespace_work.friction)]] N) ( ~s~ m) (-1) \\
&=& [[return sf_latex(namespace_work.w_f)]] J
\end{eqnarray}
$$
Total work
Adding up the work from above:
$$
\begin{eqnarray}
W_{total} &=& W_n + W_g + W_1 + W_f \\
&=& 0J + 0J + ([[return sf_latex(namespace_work.w_1)]] J) + ([[return sf_latex(namespace_work.w_f)]] J) \\
&=& [[return sf_latex(namespace_work.w_total)]] J
\end{eqnarray}
$$
(type "a" for accelerating, "d" for decelerating, or "n" for neither)
Select unit for force:
$kg$
$J$
$N$
$m/s^2$
2
Select unit for work:
$kg$
$J$
$N$
$m/s^2$
1
work || kinetic_energy
Exercise - Work and kinetic energy basic
if (int_count_times_randomized == 0){
return 8.0;
} else {
return random_min_max_precision(5, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 120;
} else {
return random_min_max_precision(100, 150, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 10;
} else {
return random_min_max_precision(-30, 30, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return -30;
} else {
return random_min_max_precision(-30, 30, -1); //defined in setup_exercise_all.js
}
return (true);
[[
namespace_work.w_total = ~w_a~ + ~w_b~;
namespace_work.ke_f = ~ke_i~ + namespace_work.w_total;
return "";
]]
An object of mass $m$ initially carries kinetic energy $KE_i$. Alice and Bob do work of $W_A$ and $W_B$ on the object respectively.
Given:
$m = ~m~ kg$.
$KE_i = ~ke_i~ J$.
$W_A = ~w_a~ J$.
$W_B = ~w_b~ J$.
Find:
Total work $W_{total}$.
The final kinetic energy $KE_f$.
Hint:
Work means energy transfer.
Positive work means the object gains energy, negative means it loses energy.
$\Delta KE = W_{total}$, where $\Delta KE = KE_f - KE_i$.
Solution
Total work
Adding up the work from above:
$$
\begin{eqnarray}
W_{total} &=& W_A + W_B \\
&=& (~w_a~ J) +( ~w_b~ J) \\
&=& [[return sf_latex(namespace_work.w_total)]] J
\end{eqnarray}
$$
If $W_{total} \gt 0J$, the block is speeding up.
If $W_{total} \lt 0J$, the block is slowing down.
In our case, $W_{total} = [[return sf_latex(namespace_work.w_total)]] J$ means the object is gaining $[[return sf_latex(Math.abs(namespace_work.w_total))]] J$ amount of $KE$.
In our case, $W_{total} = [[return sf_latex(namespace_work.w_total)]] J$ means the object is losing $[[return sf_latex(Math.abs(namespace_work.w_total))]] J$ amount of $KE$.
In our case, $W_{total} = 0 J$ means the object is neither gaining nor losing $KE$. In other words, $KE$ stays the same.
[[
namespace_work.w_total = ~w_a~ + ~w_b~;
namespace_work.ke_i = 0.5*~m~ * ~v_i~*~v_i~;
namespace_work.ke_f = namespace_work.ke_i + namespace_work.w_total;
namespace_work.v_f = Math.sqrt(2*namespace_work.ke_f/~m~);
return "";
]]
An object of mass $m$ initially moves at an initial speed $v_i$. Alice and Bob do work of $W_A$ and $W_B$ on the object respectively.
Given:
$m = ~m~ kg$.
$v_i = ~v_i~ m/s$.
$W_A = ~w_a~ J$.
$W_B = ~w_b~ J$.
Find:
Total work $W_{total}$.
The initial kinetic energy $KE_i$.
The final kinetic energy $KE_f$.
The final speed $v_f$.
Hint:
Work means energy transfer.
Positive work means the object gains energy, negative means it loses energy.
$KE = \frac{1}{2}mv^2 \Leftrightarrow v = \sqrt{\frac{2 (KE)}{m}}$.
$\Delta KE = W_{total}$, where $\Delta KE = KE_f - KE_i$.
Solution
Total work
Adding up the work from above:
$$
\begin{eqnarray}
W_{total} &=& W_A + W_B \\
&=& (~w_a~ J) +( ~w_b~ J) \\
&=& [[return sf_latex(namespace_work.w_total)]] J
\end{eqnarray}
$$
If $W_{total} \gt 0J$, the block is speeding up.
If $W_{total} \lt 0J$, the block is slowing down.
In our case, $W_{total} = [[return sf_latex(namespace_work.w_total)]] J$ means the object is gaining $[[return sf_latex(Math.abs(namespace_work.w_total))]] J$ amount of $KE$.
In our case, $W_{total} = [[return sf_latex(namespace_work.w_total)]] J$ means the object is losing $[[return sf_latex(Math.abs(namespace_work.w_total))]] J$ amount of $KE$.
In our case, $W_{total} = 0 J$ means the object is neither gaining nor losing $KE$. In other words, $KE$ stays the same.
You want to deliver a box of mass $m$ up a ramp of length $s$ at angle $\theta$. Your plan is stay at the bottom yourself, but give the box a hard push so it has an initial speed of $v_i$ and slides to the top by itself. The ramp has friction $f$ acting on the box as it slides up.
Given:
$m = ~m~ kg$.
$s = ~s~ m$.
$\theta = ~theta~ ^\circ$.
$f = [[return Math.abs(~friction~)]] N$.
Find $W_n$, the work done by the normal force.
Find $W_g$, the work done by gravity.
Find $W_f$, the work done by friction.
Find $W_{total}$, the total work on the box after your push (not including work done by your own push).
Find $KE_i$, the minimum initial kinetic energy of the box for you to succeed.
Find $v_i$, the minimum initial speed.
Hint:
$W_{total}$ tells you how much kinetic energy the box will lose on its journey to the top of the ramp.
You need to give the box enough $KE$ at the bottom to compensate for the loss.
The force diagram.
Solution
Work done by the normal force
$F_n$ and $s$.
The angle between $F_n$ and $s$ is $90^\circ$:
$$
\begin{eqnarray}
W_n &=& F_n s \cos(90^\circ) \\
&=& 0 J
\end{eqnarray}
$$
Work done by gravity
The angle between $F_g$ and $s$ is $\theta_g = 90^\circ [[return (~s~>0)? "+":"-"]] ~theta~ ^\circ = [[return sf_latex(namespace_work.theta_g)]]^\circ$.
From the figure, we see that the angle between $F_g$ and $s$ is:
$$
\begin{eqnarray}
\theta_g &=& 90^\circ [[return (~s~>0)? "+":"-"]] \theta \\
&=& 90^\circ [[return (~s~>0)? "+":"-"]] ~theta~ ^\circ \\
&=& [[return sf_latex(namespace_work.theta_g)]]^\circ
\end{eqnarray}
$$
This gives:
$$
\begin{eqnarray}
W_g &=& (mg)(s)\cos\theta_g \\
&=& (~m~kg)(9.8 m/s^2)([[return Math.abs(~s~)]] m)\cos [[return sf_latex(namespace_work.theta_g)]]^\circ \\
&=& [[return sf_latex(namespace_work.w_g)]] J
\end{eqnarray}
$$
$W_g$ and $mgh$
Overall there are two cases:
Moving up: $W_g = -mgh$, so gravity takes energy from the block as it goes up, slowing it down.
Moving down: $W_g = +mgh$, so gravity gives the block energy as it slides down, speeding it up.
We are in the [[return (~s~>0)? "first":"second"]] case in this problem. You can randomize to see the other case.
where $h= s\sin\theta$ is the vertical height of the incline.
We can see this using the trig identity $\cos (90^\circ \pm \theta) = \mp \sin \theta$:
$$
\begin{eqnarray}
W_g &=& (mg)(s)\cos\theta_g \\
&=& (mg)(s)\cos (90^\circ [[return (~s~>0)? "+":"-"]] \theta) \\
&=& mg ([[return (~s~>0)? "-":"+"]] s \sin \theta) \\
&=& [[return (~s~>0)? "-":"+"]] mgh
\end{eqnarray}
$$
Click open "Alternative approach for $W_g$" below for a more physical explanation.
We could also break displacement into two components:
$$
\begin{eqnarray}
\vec s &=& s_x \hat i + s_y \hat j \\
&=& s \cos \theta \hat i [[return (~s~>0)? "+":"-"]] s \sin\theta \hat j \\
&=& ([[return sf_latex(~s~ * Math.cos(~theta~*Math.PI/180))]] \hat i [[return (~s~>0)? "+":""]] [[return sf_latex(~s~ * Math.sin(~theta~*Math.PI/180))]] \hat j) m
\end{eqnarray}
$$
Of the two components of $\vec s$, only the vertical component will contribute to $W_g$, so we can just ignore the $\hat i$ term.
There are two equivalent ways to proceed:
Method 1: Using $W = F s cos \theta$
The vertical component of $\vec s$ points [[return (~s~>0)? "up":"down"]], therefore it is [[return (~s~>0)?"anti-parallel":"paralle"]] to $F_g$, giving an angle of $[[return (~s~>0)? "180":"0"]]^\circ$ with $F_g$:
$$
\begin{eqnarray}
W_g &=& (mg)(|s_y|) \cos [[return (~s~>0)? "180":"0"]]^\circ \\
&=& (~m~kg)(9.8 m/s^2)([[return sf_latex(Math.abs(~s~ * Math.sin(~theta~*Math.PI/180)))]] m)\cos [[return (~s~>0)? "180":"0"]]^\circ \\
&=& [[return sf_latex(namespace_work.w_g)]] J
\end{eqnarray}
$$
If we don't put in numbers, we can write $|s_y| = h = s \sin \theta$:
$$
\begin{eqnarray}
W_g &=& (mg)(|s_y|) \cos [[return (~s~>0)? "180":"0"]]^\circ \\
&=& (mg)(h) ([[return (~s~>0)? "-":"+"]] 1) \\
&=& [[return (~s~>0)? "-":"+"]] mgh = [[return (~s~>0)? "-":"+"]] mg s \sin\theta
\end{eqnarray}
$$
where $\theta$ is the angle of the incline.
Method 2: Using $W = \vec F \cdot \vec s = F_x s_x + F_y s_y$
Weight is $\vec F_g = -mg \hat j = -[[return ~m~ * 9.8]] \hat j N$ in vector form (i.e. $F_x = 0$). Taking the dot product with $\vec s= ([[return sf_latex(~s~ * Math.cos(~theta~*Math.PI/180))]] \hat i [[return (~s~>0)? "+":""]] [[return sf_latex(~s~ * Math.sin(~theta~*Math.PI/180))]] \hat j) m$ gives:
$$
\begin{eqnarray}
W_g &=& 0 + (-[[return ~m~ * 9.8]] N)([[return sf_latex(~s~ * Math.sin(~theta~*Math.PI/180))]] m) \\
&=& [[return sf_latex(namespace_work.w_g)]] J
\end{eqnarray}
$$
Or symbolically, using $\vec F_g = -mg \hat j$ and $\vec s = s_x \hat i [[return (~s~>0)? "+":"-"]] h \hat j$ (where $h = |s_y|$), we get the dot product:
$$
\begin{eqnarray}
W_g &=& 0 + (-mg)([[return (~s~>0)? "+":"-"]] h) \\
&=& [[return (~s~>0)? "-":"+"]] mgh = [[return (~s~>0)? "-":"+"]] mg s \sin\theta
\end{eqnarray}
$$
Work done by friction
$f$ and $s$.
From the figure, we see that $f$ and $s$ are [[return (~s~*~friction~>0)?"parallel":"anti-parallel"]], so the angle between them is:
$$
\begin{eqnarray}
\theta_f &=& [[return sf_latex(namespace_work.theta_t)]]^\circ
\end{eqnarray}
$$
This gives:
$$
\begin{eqnarray}
W_f &=& (f)(s)\cos\theta_f \\
&=& ([[return Math.abs(~friction~)]] N)([[return Math.abs(~s~)]] m)\cos [[return sf_latex(namespace_work.theta_t)]]^\circ \\
&=& [[return sf_latex(namespace_work.w_f)]] J
\end{eqnarray}
$$
Total work
Sum up all three works we get:
$$
\begin{eqnarray}
W_{total} &=& W_n + W_g + W_f \\
&=& 0J + ([[return sf_latex(namespace_work.w_g)]] J) + ([[return sf_latex(namespace_work.w_f)]] J) \\
&=& [[return sf_latex(namespace_work.w_total)]] J
\end{eqnarray}
$$
$W_{total} \lt 0J$ means the block slows down as it slides up the ramp, as expected.
Minimum kinetic energy
$W_{total} = [[return sf_latex(namespace_work.w_total)]] J$ means we expect the box to lose $[[return sf_latex(Math.abs(namespace_work.w_total))]] J$ of kinetic energy by the time it gets to the top of the ramp. So you must supply at least that amount at the bottom, therefore:
$$
\begin{eqnarray}
KE_i &=& -W_{total} \\
&=& [[return sf_latex(Math.abs(namespace_work.w_total))]] J
\end{eqnarray}
$$
If you do not understand, use these analogy:
You will lose 10 gallons of water (as sweat) on a long hike through a desert, how much water minimum should you bring when you set out? Answer: 10 gallons!
If you are going to spend $\$[[return sf_latex(Math.abs(namespace_work.w_total))]]$ on a European vacation, so how much money minimum should you bring when you begin your trip? Answer: $\$[[return sf_latex(Math.abs(namespace_work.w_total))]]$!
Here is an unnecessarily mathematical approach you often see in textbooks based on the work-kinetic energy theorem $W_{total} = \Delta KE = KE_f - KE_i$.
We put $KE_f=0 J$ (meaning the box is at rest when it reaches the top) to get:
$$
\begin{eqnarray}
W_{total} &=& 0- KE_i \\
\Rightarrow KE_i &=& -W_{total} \\
&=& [[return sf_latex(Math.abs(namespace_work.w_total))]] J
\end{eqnarray}
$$
My objection to this approach is that it makes something that is simple (see hiking analogy above) seems unintuitive and obscure. The "theorem" magically gives a result without the student's understanding.
if (int_count_times_randomized == 0){
return 100;
} else {
return random_min_max_precision(50, 200, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 8;
} else {
return random_min_max_precision(-15, 15, 0, false, -5, 5); //defined in setup_exercise_all.js
}
return (true);
A mass tied to a spring. The equilibrium position of the spring's right tip is also indicated.
[[
namespace_work.f_s = - ~k~ * ~x~ * 0.01;
namespace_work.f_you = -namespace_work.f_s;
return "";
]]
A mass is tied to a spring with spring constant $k = ~k~ N/m$. The spring is [[return (~x~>0)? "stretched":"compressed"]] by $[[return Math.abs(~x~)]] cm$.
Find the force $F_s$ exerted on the mass by the spring.
Find the force $F_{you}$ you have to exert on the mass to hold the mass in place.
Enter a positive number for a force pointing to the right, negative for one pointing left.
Hint: $F_s = -kx$.
The red arrow is $F_s$, which is always opposite to $x$.
Solution
Since the spring is [[return (~x~>0)? "stretched":"compressed"]], we have $x = ~x~ cm$.
$$
\begin{eqnarray}
F_s &=& - k x \\
&=& -(~k~ N/m) (~x~ \times 0.01 m) \\
&=& [[return sf_latex(namespace_work.f_s)]] N
\end{eqnarray}
$$
$F_s$ being [[return (~x~>0)? "negative":"positive"]] means the spring is [[return (~x~>0)? "pulling the mass to the left.":"pushing the mass to the right."]]
To hold the mass in place, you just apply a force opposite to $F_s$:
$$
F_{you} = - F_s = [[return sf_latex(namespace_work.f_you)]] N
$$
$F_s= $
return sf_math(namespace_work.f_s)
5%
$F_{you}= $
return sf_math(namespace_work.f_you)
5%
Select unit for force:
$m$
$Nm$
$J$
$N$
3
work || force || hooke || spring
Exercise - Work done by a spring
if (int_count_times_randomized == 0){
return 100;
} else {
return random_min_max_precision(50, 200, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 8;
} else {
return random_min_max_precision(-15, 15, 0, false, -5, 5); //defined in setup_exercise_all.js
}
return (true);
A mass tied to a spring. The equilibrium position of the spring's right tip is also indicated.
[[
namespace_work.f_s = - ~k~ * ~x~ * 0.01;
namespace_work.f_you = -namespace_work.f_s;
namespace_work.w_s = -0.5 * ~k~ * ~x~ * 0.01 * ~x~ * 0.01;
namespace_work.w_you = - namespace_work.w_s;
return "";
]]
A mass is tied to a spring with spring constant $k = ~k~ N/m$. Starting from the equilibrium position, you apply a force $F_{you}$ to slowly [[return (~x~>0)? "stretch":"compress"]] by $[[return Math.abs(~x~)]] cm$.
Find the work $W_s$ done by the spring.
Find the work $W_{you}$ done by you.
Hint:
$F_s = -kx$.
$W = \int \vec F \cdot d \vec s = \int F_x dx$.
The red arrow is $F_s$, which is always opposite to $x$.
To move the block "slowly" (physicist's way to say "without noticable acceleration"), $F_{you}$ must be opposite to $F_s$ and equal in magnitude to cancel it out. Therefore $F_{you} = -F_s = +kx$:
$$
\begin{eqnarray}
W_{you} &=& \int_0^x F_{you} dx \\
&=& \int_0^x (+kx) dx \\
&=& +k \int_0^x x dx \\
&=& +\frac{1}{2} k x^2 \\
&=& +\frac{1}{2} (~k~ N/m) (~x~ \times 0.01m)^2 \\
&=& [[return sf_latex(namespace_work.w_you)]] J
\end{eqnarray}
$$
$W_s= $
return sf_math(namespace_work.w_s)
5%
$W_{you}= $
return sf_math(namespace_work.w_you)
5%
Select unit for work:
$m$
$Nm$
$J$
$N$
2
energy || work || power
Exercise - Power basic
if (int_count_times_randomized == 0){
return 100;
} else {
return random_min_max_precision(60, 150, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 0.40;
} else {
return random_min_max_precision(0.25, 0.5, 2); //defined in setup_exercise_all.js
}
return (true);
[[
namespace_work.t = ~t_in_hour~ * 3600;
namespace_work.energy = ~p~ * namespace_work.t;
namespace_work.energy_in_kwh = ~p~ / 1000 * ~t_in_hour~;
namespace_work.cost = namespace_work.energy_in_kwh * ~cost_per_kwh~;
return "";
]]
A light bulb has a power rating of $~p~ W$, which you turn on for $~t_in_hour~$ hours. The power company charges $ \$[[return (~cost_per_kwh~).toFixed(2)]]$ per $kWh$.
Find $E$, the energy it consumes in $J$ and in $kWh$.
How much money ($M$) do you owe the power company?
Very big or very small numbers can be entered as "1.234e-6" (for $ 1.234\times 10^{-6}$).
Hint:
$E = W = P t$: energy consumed by the light bulb.
$1W = 0.001kW$, and $1kW \times 1h = 1kWh$.
Solution
$P = ~p~ W$ means the light bulb consumes $~p~ J$ every second. Therefore multiplying it with the time will gives the total energy consumed.
Energy in SI unit
$$
\begin{eqnarray}
E &=& P t \\
&=& (~p~ W)(~t_in_hour~ h) \\
&=& (~p~ J/s)(~t_in_hour~ \times 3600 s) \\
&=& [[return sf_latex(namespace_work.energy)]] J
\end{eqnarray}
$$
if (int_count_times_randomized == 0){
return 10;
} else {
return random_min_max_precision(5, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 20;
} else {
return random_min_max_precision(3, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 7;
} else {
return random_min_max_precision(5, 20, 0); //defined in setup_exercise_all.js
}
return (true);
Moving a mass from $A$ to $B$ via two different paths. $B$ is vertically above $P$. Not drawn to scale.
An object is moved from $A (0, 0)$ to $B (~X~, ~Y~)$ along two different paths. Show your work explicitly using $W = F s \cos \theta$ for the constant force in this problem, so we can see the difference in how you calculate the work in the two cases below.
Calculate the work done by gravity via Path 1.
Calculate the work done by gravity via Path 2 (i.e. passing through $P$).
Hint: Be careful with the angle $\theta$ for Path 1. The angle should be greater than 90 degrees.
The angle between the weight and displacement for Path 1 is $\theta + 90^\circ$. $s$ is the total length of Path 1.
Path 1
The angle between the force of gravity and the displacement is $\theta + 90^\circ$ for Path 1. Therefore the work done by gravity is:
$$
\begin{eqnarray}
W_1 &=& (mg)(s) \cos (\theta + 90^\circ) \\
&=& (mg)(s)(-\sin \theta) \qquad \text{Trig identity} \\
&=& -mg s \sin\theta \\
&=& -mg Y \qquad \text{ because } s \sin\theta = Y \\
&=& - (~m~ kg)(9.8 m/s^2) (~Y~ m) \\
&=& [[return sf_latex(W_1)]] J
\end{eqnarray}
$$
Path 2
The angle between the force of gravity and the displacement is $90^\circ$ for Path 2 from $A$ to $P$. Therefore the work done by gravity is:
$$
\begin{eqnarray}
W_2 (A\rightarrow P) &=& (mg)(X) \cos 90^\circ \\
&=& 0 J
\end{eqnarray}
$$
The angle between the force of gravity and the displacement is $180^\circ$ for Path 2 from $P$ to $B$. Therefore the work done by gravity is:
$$
\begin{eqnarray}
W_2 (P\rightarrow B) &=& (mg)(Y) \cos (180^\circ) \\
&=& (mg)(Y)(-1) \\
&=& -mg Y \\
&=& - (~m~ kg)(9.8 m/s^2) (~Y~ m) \\
&=& [[return sf_latex(W_2)]] J
\end{eqnarray}
$$
Therefore:
$$
\begin{eqnarray}
W_2 &=& W_2 (A\rightarrow P) + W_2 (P\rightarrow B) \\
&=& 0 + [[return sf_latex(W_2)]] J \\
&=& [[return sf_latex(W_2)]] J
\end{eqnarray}
$$
Comparison
The work done by gravity is the same for the two paths (i.e. $W_1 = W_2$). The force of gravity is conservative.
$W_1 = $
return sf_math(W_1)
5%
$J$ $W_2 = $
return sf_math(W_2)
5%
$J$
Is the force of gravity conservative?
Yes
No
0
electricity || e_field || e_potential
Example - Path dependence of work 2
Two paths to travel from $A$ to $B$, passing through $P$ and $Q$ respectively.
The figure shows two paths to travel from $A (0, 0)$ to $B (X, Y)$. Check if $W = \int_A^B \vec F \cdot d\vec s$ gives the same answer for the two paths in the following cases:
$\vec F = \hat i$.
$\vec F = x \hat i$.
$\vec F = y \hat i$.
$\vec F = y \hat i + x \hat j$.
Solution
$\vec F = \hat i$
Path 1:
$$
\begin{eqnarray}
W_1 &=& \int_{\text{path 1}} \vec F \cdot d\vec s \\
&=& \int_A^P \vec F \cdot d\vec s + \int_P^B \vec F \cdot d\vec s \\
&=& \int_A^P (\hat i) \cdot (dy \hat j) + \int_P^B (\hat i) \cdot (dx \hat i) \\
&=& 0 + \int_0^X dx \\
&=& X
\end{eqnarray}
$$
Path 2:
$$
\begin{eqnarray}
W_2 &=& \int_{\text{path 2}} \vec F \cdot d\vec s \\
&=& \int_A^Q \vec F \cdot d\vec s + \int_Q^B \vec F \cdot d\vec s \\
&=& \int_A^Q (\hat i) \cdot (dx \hat i) + \int_Q^B (\hat i) \cdot (dy \hat j) \\
&=& \int_0^X dx + 0 \\
&=& X
\end{eqnarray}
$$
Therefore $W_1 = W_2$, and the $F$ is conservative. $U= -x$ can generate the force.
$\vec F = x \hat i$
Path 1:
$$
\begin{eqnarray}
W_1 &=& \int_{\text{path 1}} \vec F \cdot d\vec s \\
&=& \int_A^P \vec F \cdot d\vec s + \int_P^B \vec F \cdot d\vec s \\
&=& \int_A^P (x\hat i) \cdot (dy \hat j) + \int_P^B (x\hat i) \cdot (dx \hat i) \\
&=& 0 + \int_0^X x dx \\
&=& \frac{1}{2}X^2
\end{eqnarray}
$$
Path 2:
$$
\begin{eqnarray}
W_2 &=& \int_{\text{path 2}} \vec F \cdot d\vec s \\
&=& \int_A^Q \vec F \cdot d\vec s + \int_Q^B \vec F \cdot d\vec s \\
&=& \int_A^Q (x \hat i) \cdot (dx \hat i) + \int_Q^B (x \hat i) \cdot (dy \hat j) \\
&=& \int_0^X x dx + 0 \\
&=& \frac{1}{2}X^2
\end{eqnarray}
$$
Therefore $W_1 = W_2$, and the $F$ is conservative. $U= -\frac{1}{2}x^2$ can generate the force.
$\vec F = y \hat i$
Path 1:
$$
\begin{eqnarray}
W_1 &=& \int_{\text{path 1}} \vec F \cdot d\vec s \\
&=& \int_A^P \vec F \cdot d\vec s + \int_P^B \vec F \cdot d\vec s \\
&=& \int_A^P (y \hat i) \cdot (dy \hat j) + \int_P^B (y \hat i) \cdot (dx \hat i) \\
&=& 0 + \int_P^B (Y \hat i) \cdot (dx \hat i) \qquad \text{note that $y=Y$ along $P$ to $B$.}\\
&=& 0 + \int_0^X Y dx\\
&=& XY
\end{eqnarray}
$$
Path 2:
$$
\begin{eqnarray}
W_2 &=& \int_{\text{path 2}} \vec F \cdot d\vec s \\
&=& \int_A^Q \vec F \cdot d\vec s + \int_Q^B \vec F \cdot d\vec s \\
&=& \int_A^Q (y \hat i) \cdot (dx \hat i) + \int_Q^B (y \hat i) \cdot (dy \hat j) \\
&=& \int_A^Q (0 \hat i) \cdot (dx \hat i) + 0 \qquad \text{note that $y=0$ along $A$ to $Q$.} \\
&=& 0 + 0 \\
&=& 0
\end{eqnarray}
$$
Therefore $W_1 \neq W_2$, and the $F$ is non-conservative. There is no $U$ that can generate the force via $\vec F = -\nabla U$.
$\vec F = y \hat i + x \hat j$
Path 1:
$$
\begin{eqnarray}
W_1 &=& \int_{\text{path 1}} \vec F \cdot d\vec s \\
&=& \int_A^P \vec F \cdot d\vec s + \int_P^B \vec F \cdot d\vec s \\
&=& \int_A^P (y \hat i + x \hat j) \cdot (dy \hat j) + \int_P^B (y \hat i + x \hat j) \cdot (dx \hat i) \\
&=& \int_A^P (y \hat i + 0 \hat j) \cdot (dy \hat j) + \int_P^B (Y \hat i + x \hat j) \cdot (dx \hat i) \\
&=& 0 + \int_0^X Y dx \\
&=& XY
\end{eqnarray}
$$
Path 2:
$$
\begin{eqnarray}
W_2 &=& \int_{\text{path 2}} \vec F \cdot d\vec s \\
&=& \int_A^Q \vec F \cdot d\vec s + \int_Q^B \vec F \cdot d\vec s \\
&=& \int_A^Q (y \hat i + x \hat j) \cdot (dx \hat i) + \int_Q^B (y \hat i + x \hat j) \cdot (dy \hat j) \\
&=& \int_A^Q (0 \hat i + x \hat j) \cdot (dx \hat i) + \int_Q^B (y \hat i + X \hat j) \cdot (dy \hat j) \\
&=& 0 + \int_0^Y X dy \\
&=& XY
\end{eqnarray}
$$
Therefore $W_1 = W_2$, and the $F$ is conservative. $U= -xy$ can generate the force.
energy || work
Exercise - Work and force to stop a car
if (int_count_times_randomized == 0){
return 1000;
} else {
return random_min_max_precision(700, 1500, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 30;
} else {
return random_min_max_precision(5, 40, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(0.5, 5, 1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 110;
} else {
return random_min_max_precision(50, 200, 0); //defined in setup_exercise_all.js
}
return (true);
A car of mass $m = ~m~ kg$ is moving at $v_0 = ~v_0~ m/s$. The car is brought to a stop by a force.
What is the initial kinetic energy of the car?
Calculate the total work required to stop the car.
Use work to find the magnitude of a constant force $F_1$ that can stop the car a distance of $x_1 = ~x_1~ m$.
Use the kinematic equations to find the acceleration of the car during the stop.
Use Newton's second law to calculate the magnitude of the force.
Very big or very small numbers can be entered as "1.234e-6" (for $ 1.234\times 10^{-6}$).
Hint: $W = \vec F \cdot \vec s$
Kinematic equations:
$$
\begin{eqnarray}
v &=& v_0 + at \tag{$s$} \\
s - s_0 &=& v_0t + \frac{1}{2} a t^2 \tag{$v$}\\
s - s_0 &=& \frac{1}{2} (v_0+v)t \tag{$a$} \\
v^2 &=& v_0^2 + 2a(s-s_0) \tag{$t$}
\end{eqnarray}
$$
The initial kinetic energy of the car is:
$$ KE_i = \frac{1}{2} m v_0^2 = \frac{1}{2} (~m~) (~v_0~)^2 = [[return sf_latex(KE_i)]] J $$
Work
The work done by the force is equal to the change in kinetic energy of the car. The initial kinetic energy of the car is:
$$ W = KE_f - KE_i = 0 - KE_i = [[return sf_latex(W)]] J $$
Force
The magnitude of the force required to stop the car a distance of $x_1 = ~x_1~ m$ is:
$$
\begin{eqnarray}
W &=& F_1 x_1 \cos 180^\circ = -F_1 x_1 \\
\Rightarrow F_1 &=& -\frac{W}{x_1} \\
&=& -\frac{[[return sf_latex(W)]] J}{~x_1~ m} \\
&=& [[return sf_latex(F_1)]] N
\end{eqnarray}
$$
Acceleration
$s_0 = 0m$
$v_0 = ~v_0~ m/s$
$s = ~x_1~ m$
$v = 0 m/s$
$a = ?$
$t = $ missing
The acceleration of the car during the stop can be found from the $t$-equation of the kinematic equations:
$$
\begin{eqnarray}
v^2 &=& v_0^2 + 2 a x_1 \\
0 &=& v_0^2 + 2 a x_1 \\
\Rightarrow a &=& -\frac{v_0^2}{2 x_1} = -\frac{(~v_0~)^2}{2 \times ~x_1~} \\
&=& [[return sf_latex(a)]] m/s^2
\end{eqnarray}
$$
Force
By Newton's second law, the magnitude of the force required to stop the car is:
$$
\begin{eqnarray}
F_1 &=& m a \\
&=& (~m~ kg)(|[[return sf_latex(a)]] m/s^2|) \qquad \text{absolute value because we are asked to find only the magnitude} \\
&=& [[return sf_latex(F_1)]] N
\end{eqnarray}
$$
The force is the same whether it is calculated from work or from Newton's second law.
$KE_i = $
return sf_math(KE_i)
5%
$J$ $W = $
return sf_math(W)
5%
$J$ Force computed by work =
return sf_math(F_1)
5%
$N$ $a = $
return sf_math(a)
5%
$m/s^2$ Force computed from the second law =
return sf_math(F_1)
5%
$N$
energy || friction
Exercise - Distance to stop a block with friction
if (int_count_times_randomized == 0){
// Default: 5 kg
return 5;
} else {
// Randomize mass between 1 and 20 kg, integer
return random_min_max_precision(1, 20, 0);
}
if (int_count_times_randomized == 0){
// Default: 8 m/s
return 8;
} else {
// Randomize velocity between 2 and 25 m/s, integer
return random_min_max_precision(2, 25, 0);
}
if (int_count_times_randomized == 0){
// Default: 0.30
return 0.30;
} else {
// Randomize friction coefficient between 0.10 and 0.80, with 2 decimal places
return random_min_max_precision(0.10, 0.80, 2);
}
return true;
A block of mass $~m~ \text{ kg}$ is sliding on a horizontal surface.
The kinetic friction coefficient between the block and the surface is $~mu~$.
Initially, the block moves at $~v0~ \text{ m/s}$.
Calculate the initial kinetic energy.
Calculate the magnitude of the force of friction.
Using energy considerations, find the distance $d$ the block travels before coming to rest
Hint:
Since the normal force in this case is $F_n = mg$, friction is $f = \mu F_n = \mu mg$,
and the work done by friction is $W = f d \cos 180^\circ$.
/*
Using conservation of energy with friction:
Initial kinetic energy: KE_i = (1/2) * m * v0^2
Work done by friction: W_f = f * d = (mu * m * g) * d
Energy balance: KE_i - W_f = 0 => (1/2) m v0^2 = mu m g d
=> d = [ (1/2) m v0^2 ] / [ mu m g ] = (v0^2) / (2 mu g)
*/
g = 9.8; // gravitational acceleration (m/s^2)
KE_i = (1/2) * ~m~ * ~v0~ * ~v0~;
friction = ~mu~ * ~m~ * g;
distance = (~v0~ * ~v0~) / (2 * ~mu~ * g);
Solution
The initial kinetic energy is
$$
KE_\text{initial} = \frac{1}{2} m v_0^2 = \frac{1}{2} \, (~m~) \, (~v0~)^2 = [[return sf_latex(KE_i)]] \, \mathrm{J}.
$$
Since there are only two vertical forces, normal force $F_n$ (up) and gravity $mg$ (down). The lack of vertical motion means these two vertical forces cancel, therefore the normal force is $F_n = mg$.
The friction is:
$$
f = \mu F_n= \mu mg = (~mu~) (~m~) (9.8) = [[return sf_latex(friction)]] \, \mathrm{N}.
$$
The work done by friction over distance $d$ is
$$
W_f = f d \cos 180^\circ = -\mu m g d.
$$
By work-energy theorem, for the block to come to rest (i.e. $KE_\text{final} = 0 \, \mathrm{J}$):
$$
\begin{eqnarray}
W_f &=& \Delta KE = KE_\text{final} - KE_\text{initial} = 0 - KE_\text{initial} \\
\Rightarrow -\mu m g d &=& - KE_\text{initial} \\
\Rightarrow d &=& \frac{KE_\text{initial}}{\mu m g} \\
&=& \frac{\frac{1}{2} m v_0^2}{\mu m g} = \frac{v_0^2}{2 \mu g} \\
&=& \frac{(~v0~)^2}{2 (~mu~) (9.8)} \\
&=& [[return sf_latex(distance)]] \, \mathrm{m}
\end{eqnarray}
$$
A force acting on a particle along the $x$-axis varies with position according to the equation:
$$ F(x) = ~a~x^2 $$
where $F$ is in newtons and $x$ is in meters. Calculate the total work done by this force as the particle moves from $x = ~x_1~\,\mathrm{m}$ to $x = ~x_2~\,\mathrm{m}$.
Hint:
The work done by a variable force $F(x)$ over a displacement from $x_1$ to $x_2$ is given by the integral:
$$ W = \int_{x_1}^{x_2} F(x)\, dx $$
In this case, integrate $F(x) = ~a~x^2$ from $x = ~x_1~$ to $x = ~x_2~$.
a = ~a~;
x1 = ~x_1~;
x2 = ~x_2~;
W = a * (Math.pow(x2, 3) - Math.pow(x1, 3)) / 3;
Solution
The work done by the force is the area under the force-position graph, calculated via:
$$
\begin{eqnarray}
W &=& \int_{x_1}^{x_2} F(x)\, dx = \int_{~x_1~}^{~x_2~} ~a~x^2\, dx \\
&=& ~a~ \int_{~x_1~}^{~x_2~} x^2\, dx = ~a~ \left[\frac{x^3}{3}\right]_{~x_1~}^{~x_2~} \\
&=& ~a~ \left(\frac{~x_2~^3 - ~x_1~^3}{3}\right) \\
&=& [[return sf_latex(W)]]\,\mathrm{J}
\end{eqnarray}
$$
