What mistake am I making when calculating work done by a force?

In summary: In this example, displacement is positive because the force is positive and the x coordinate decreases. So the work done is negative.
  • #1
songoku
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Homework Statement
Draw the graph of ##F=-2x## and calculate the work done from ##[-1,1]##
Relevant Equations
Work = area under the graph
The answer key is zero because the areas are above and below x-axis and have equal magnitude so canceling out each other.

But I am confused about the solution
1645842548574.png


Area 1 is above x-axis but I think the work done is negative since the sign of ##F## and ##x## is opposite. Work done on area 2 is also negative for the same reasoning so my answer would be -2 J

Where is my mistake? Thanks
 
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  • #2
songoku said:
Homework Statement:: Draw the graph of ##F=-2x## and calculate the work done from ##[-1,1]##
Relevant Equations:: Work = area under the graph

The answer key is zero because the areas are above and below x-axis and have equal magnitude so canceling out each other.

But I am confused about the solution
View attachment 297599

Area 1 is above x-axis but I think the work done is negative since the sign of ##F## and ##x## is opposite. Work done on area 2 is also negative for the same reasoning so my answer would be -2 J

Where is my mistake? Thanks
Ask yourself, "What is the definition of Work?"
 
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  • #3
SammyS said:
Ask yourself, "What is the definition of Work?"
Work is the product of displacement and force which is parallel to the displacement

So based on that definition, I think when the force is positive and displacement is negative, the work done will be negative.

Thanks
 
  • #4
songoku said:
Work is the product of displacement and force which is parallel to the displacement

So based on that definition, I think when the force is positive and displacement is negative, the work done will be negative.

Thanks
Right. Looks like you've got it.

So, in going from ##x=-1## to ##x=0##, for instance, the displacement is positive, even though ##x## is negative.
 
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  • #5
For a non constant force the work in going from1 to 2 is $$work=\int _1^2 F(x) \,
dx$$ The increment dx is positive in the +x direction. So the net result result
$$work=\int _{x=-1}^{x=+1} F(x) \,
dx=0$$
is zero
 
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  • #6
SammyS said:
Right. Looks like you've got it.

So, in going from ##x=-1## to ##x=0##, for instance, the displacement is positive, even though ##x## is negative.
Ah I see, so one work is indeed positive (area 1) and the other one is negative (area 2) so the total is zero.

Thank you very much SammyS and hutchphd
 
  • #7
Put simply: Displacement is the change in x, not x itself.
 
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Related to What mistake am I making when calculating work done by a force?

1. What is work done by a force?

Work done by a force is a physical quantity that measures the amount of energy transferred to an object when a force is applied to it. It is defined as the product of the force applied and the displacement of the object in the direction of the force.

2. How is work done by a force calculated?

The work done by a force can be calculated by multiplying the magnitude of the force by the distance the object moves in the direction of the force. This can be expressed as W = F x d, where W is the work done, F is the force applied, and d is the displacement of the object.

3. What are the units of work done by a force?

The units of work done by a force are joules (J) in the International System of Units (SI). In the English system, the unit of work is foot-pound (ft-lb). Both units represent the amount of energy transferred to an object.

4. How does the angle between the force and displacement affect the work done?

The angle between the force and displacement affects the work done by changing the effective force applied in the direction of displacement. When the force and displacement are parallel (angle = 0°), the work done is maximum. When they are perpendicular (angle = 90°), the work done is zero. The work done is negative when the angle is obtuse (between 90° and 180°) and positive when the angle is acute (between 0° and 90°).

5. Can work done by a force be negative?

Yes, work done by a force can be negative. This happens when the force is applied in the opposite direction of the displacement. In this case, the force is doing negative work, which means it is removing energy from the object instead of transferring it. For example, when a person pushes a box up a ramp, the force of gravity is doing negative work on the box as it moves downward.

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