Work and the Frictional Force (Grade 12 Physics)

Click For Summary
SUMMARY

The discussion centers on a Grade 12 physics problem involving a boy pulling an empty wagon with a force of 111 N at an angle of 31° below the horizontal, while a frictional force of 155 N opposes the motion. The total work done by the boy is 465 J, leading to a calculated distance of 4.89 m. Participants clarify that the frictional force does not affect the distance calculation directly, as the work done is based on the applied force and angle, despite the friction slowing the wagon down. The net work done is crucial for understanding the dynamics of the situation.

PREREQUISITES
  • Understanding of work-energy principles in physics
  • Familiarity with vector components of forces
  • Knowledge of the work formula: W = Fdcosθ
  • Concept of net force and its impact on motion
NEXT STEPS
  • Study the concept of net work and its calculation in physics problems
  • Learn about the effects of friction on motion and energy loss
  • Explore vector decomposition of forces in inclined plane scenarios
  • Investigate the relationship between work, energy, and displacement in physics
USEFUL FOR

This discussion is beneficial for high school physics students, educators teaching mechanics, and anyone interested in understanding the principles of work and friction in motion scenarios.

EmilyBergendahl
Messages
32
Reaction score
4

Homework Statement


A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31° below the horizontal]. A frictional force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is _____.

Homework Equations


W = Fdcosθ

The Attempt at a Solution


465 J = (111 N)d(cos31°)
d = (465 J)/(111 N)(cos31°)
d = 4.89 m

Where does the frictional force come in here? Does it actually affect the distance traveled or does it just slow down the wagon?
 
Physics news on Phys.org
EmilyBergendahl said:
Where does the frictional force come in here? Does it actually affect the distance traveled or does it just slow down the wagon?

you are given the total work done in moving a distance say d on the incline-
so write down the total work done as force times the distance - but this force should be net force...
the wagon must be overcoming the frictional force as well as gravitational pull so correct your work done expression.
 
  • Like
Likes   Reactions: EmilyBergendahl
I'm still confused...

Do I need to look at force from both a vertical and horizontal component to solve the problem? Many thanks for the response!
 
EmilyBergendahl said:
I'm still confused...

Do I need to look at force from both a vertical and horizontal component to solve the problem? Many thanks for the response!
Is the wagon moving along level ground or along an incline? From the wording of the problem, I imagine the wagon is moving along level ground and the boy is pulling it at an angle 31° below the horizontal level. Is that right?
 
  • Like
Likes   Reactions: EmilyBergendahl
I believe that that is the case.

So if the wagon is moving along level ground, would I calculate the net force using the horizontal component of the applied force and the frictional force? And then input into the formula for work?
 
EmilyBergendahl said:
I believe that that is the case.

So if the wagon is moving along level ground, would I calculate the net force using the horizontal component of the applied force and the frictional force? And then input into the formula for work?
Yes.
 
  • Like
Likes   Reactions: EmilyBergendahl
Great! Thank you so much for your help! :)
 
Upon calculating, I got a negative value for distance (-7.77 m). For some reason, this doesn't make sense to me. It seems with the wording of the question that the boy is moving forwards, but I may be mistaken.
 
EmilyBergendahl said:
Upon calculating, I got a negative value for distance (-7.77 m). For some reason, this doesn't make sense to me. It seems with the wording of the question that the boy is moving forwards, but I may be mistaken.
The net force is coming out to be negative, which means there is retardation of the body. This means the body is losing energy. This makes the work done on the body negative. So, negative work and negative force will give you a positive displacement.
 
  • Like
Likes   Reactions: EmilyBergendahl
  • #10
cnh1995 said:
The net force is coming out to be negative, which means there is retardation of the body. This means the body is losing energy. This makes the work done on the body negative. So, negative work and negative force will give you a positive displacement.

So, I would use the value of -465 J when calculating the distance due to the fact that the work is acting in the same direction as the net force?
 
  • #11
EmilyBergendahl said:
work is acting in the same direction as the net force?
Work does not act on a body. It is a scalar quantity. Displacement and force are in the opposite directions, which gives a negative work.
 
  • Like
Likes   Reactions: EmilyBergendahl
  • #12
cnh1995 said:
Work does not act on a body. It is a scalar quantity. Displacement and force are in the opposite directions, which gives a negative work.

Okay, that makes sense.

So...
W = Fd
-465 J = (-59.9 N)d
d = (-465 J)/(-59.9 N)
d = 7.61 m
 
  • #13
EmilyBergendahl said:

Homework Statement


A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31° below the horizontal]. A frictional force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is _____.

Homework Equations


W = Fdcosθ

The Attempt at a Solution


465 J = (111 N)d(cos31°)
d = (465 J)/(111 N)(cos31°)
d = 4.89 m

Where does the frictional force come in here? Does it actually affect the distance traveled or does it just slow down the wagon?
The force of friction does not matter when you calculate the displacement. Given the force the boy exerts, given the angle between force and displacement, and given the work of the boy. The boy's force and work are independent of the force and work of friction, the friction slows down the wagon.
 
  • #14
EmilyBergendahl said:
I believe that that is the case.

So if the wagon is moving along level ground, would I calculate the net force using the horizontal component of the applied force and the frictional force? And then input into the formula for work?
That would be the net work done on the wagon. Why is it relevant when you are given the force the boy exerts, the work the boy does and the question is the distance traveled during the work is done?
 
  • #15
ehild said:
That would be the net work done on the wagon. Why is it relevant when you are given the force the boy exerts, the work the boy does and the question is the distance traveled during the work is done?

Okay. So I should ignore the frictional force all together in this instance?
 
  • #16
EmilyBergendahl said:
Okay. So I should ignore the frictional force all together in this instance?
Yes. Your first solution was correct.
 
  • #17
ehild said:
Yes.

Okay! Thank you so much for your clarification, ehild! :smile:
 
  • #18
You are welcome :smile:
 
  • #19
drvrm said:
you are given the total work done in moving a distance say d on the incline-
so write down the total work done as force times the distance - but this force should be net force...
the wagon must be overcoming the frictional force as well as gravitational pull so correct your work done expression.
You are wrong. The work of the boy is given, read the problem text more carefully. Also, the wagon moves on level ground, gravitation is irrelevant.
 
  • #20
ehild said:
You are wrong. The work of the boy is given, read the problem text more carefully. Also, the wagon moves on level ground, gravitation is irrelevant.

thanks for pointing the error- perhaps hurriedly i missed the point of horizontal displacement.. certainly .i should be more careful.
 
  • #21
EmilyBergendahl said:
A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31° below the horizontal]. A frictional force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is _____.

ehild said:
ou are wrong. The work of the boy is given, read the problem text more carefully. Also, the wagon moves on level ground, gravitation is irrelevant.

sorry to interject;
i again saw the wagon and the boy problem and my visualization is the following-

!. A wagon is moving with some velocity V and
2. a boy is pulling the wagon perhaps to stop it at certain angle below the horizontal through a rope or such thing.
3. we have to calculate the distance traveled by the wagon such as it stops( though the final velocity to be zero is not explicitly written.
We are given with :
The work done by the boy -
the angle at which the boy is pulling-
the pulling force by the boy -
the frictional force on the wagon-

so two types of work is leading the wagon to stop.
A the work done by the boy and
B the work done by the frictional force

Am I correct in the above visualization?
 
  • #22
drvrm said:
so two types of work is leading the wagon to stop.
A the work done by the boy and
B the work done by the frictional force

Am I correct in the above visualization?
Is there any mention in the problem that the wagon stops?
If you want to determine the distance the wagon travels before stopping, how would you get it?
 
  • #23
ehild said:
Is there any mention in the problem that the wagon stops?

yes you are right ...at the place where it should be there is a gap , that's why i said its not explicitly mentioned.
the problem may be modified say ...(till its velocity is 75% or so of the initial... ) to fill in the blanck.
then it looks better.
 
  • #24
I admit that the text of the problem is confusing, may be, on purpose. The student must clearly distinguish between the work done on an object by a certain force, and the net work done on an object. And the student must know the definition of the work of a certain force: it is the scalar product of the force with displacement of the point of attack of the force.
Even if it was meant that the wagon stopped during the time the boy pulled it and did the given work of 465 J, the distance traveled would be the same.
 
  • Like
Likes   Reactions: drvrm

Similar threads

  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
7
Views
2K
Replies
18
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K