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Work and the Frictional Force (Grade 12 Physics)

  • #1

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 travelled or does it just slow down the wagon?
 

Answers and Replies

  • #2
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Where does the frictional force come in here? Does it actually affect the distance travelled 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.
 
  • #3
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!
 
  • #4
cnh1995
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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?
 
  • #5
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?
 
  • #6
cnh1995
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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.
 
  • #7
Great! Thank you so much for your help! :)
 
  • #8
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.
 
  • #9
cnh1995
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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.
 
  • #10
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
cnh1995
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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.
 
  • #12
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
ehild
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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
ehild
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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
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
ehild
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Okay. So I should ignore the frictional force all together in this instance?
Yes. Your first solution was correct.
 
  • #17
Yes.
Okay! Thank you so much for your clarification, ehild! :smile:
 
  • #18
ehild
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You are welcome :smile:
 
  • #19
ehild
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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
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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
963
213
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 _____.
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 travelled 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
ehild
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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
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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 , thats 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
ehild
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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.
 

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