# Work and the Frictional Force (Grade 12 Physics)

• EmilyBergendahl
In summary: Yes.Great! Thank you so much for your help! :)In summary, the boy exerts a force of 111 N downward and the wagon has a frictional force of 155 N which is slowing it down. The distance traveled is 4.89 m.
EmilyBergendahl

## 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 _____.

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?

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.

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?

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.

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.

EmilyBergendahl
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?

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.

EmilyBergendahl
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

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 _____.

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.

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?

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?

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

ehild said:
Yes.

Okay! Thank you so much for your clarification, ehild!

You are welcome

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.

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.

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?

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?

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.

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.

drvrm

## 1. What is work in terms of physics?

Work in physics refers to the transfer of energy that occurs when a force is applied to an object and the object is displaced in the direction of the force. It is measured in joules (J) and is calculated by multiplying the force applied by the distance the object is moved.

## 2. How is work related to the frictional force?

The frictional force is a type of force that opposes the motion of an object and is caused by the interaction between two surfaces. When an object is moving against a surface with friction, work is being done to overcome this force and move the object. The amount of work done is directly proportional to the amount of frictional force present.

## 3. Can the frictional force ever do positive work?

No, the frictional force can never do positive work because it always acts in the opposite direction of the motion of the object. This means that it is always causing a decrease in the kinetic energy of the object and therefore doing negative work.

## 4. How does the coefficient of friction affect the amount of work done against friction?

The coefficient of friction is a measure of the roughness or smoothness of the surfaces in contact. A higher coefficient of friction means there is a greater resistance to motion and therefore more work needs to be done to overcome this force. So, a higher coefficient of friction will result in more work being done against the frictional force.

## 5. Can the frictional force be completely eliminated?

No, the frictional force cannot be completely eliminated as it is a natural force that occurs whenever two surfaces are in contact. However, it can be reduced by using lubricants or smoother surfaces, but it can never be completely eliminated.

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