How can "work" done by a force be negative?

[Manish_529]

Homework Statement

How can "work" done by a force be negative?

Relevant Equations

W=F.S (work done by a force is the dot product of the force and the displacement of the particle)

If the work done by a force is the consequence of the dot product of the force and the displacement of the particle, then it should be a scalar quantity which is true as well. But then why does question of the work done being positive or negative (depending on the angle between the force vector and the displacement vector of the particle) even arise. Shouldn't we just be concerned only about it's magnitude like we do in the case of other scalar quantities like speed or K.E or distance??

 

[PeroK]

Homework Statement: How can "work" done by a force be negative?
Relevant Equations: W=F.S (work done by a force is the dot product of the force and the displacement of the particle)

If the work done by a force is the consequence of the dot product of the force and the displacement of the particle, then it should be a scalar quantity which is true as well. But then why does question of the work done being positive or negative (depending on the angle between the force vector and the displacement vector of the particle) even arise. Shouldn't we just be concerned only about it's magnitude like we do in the case of other scalar quantities like speed or K.E or distance??

I never fully understand these questions, because the thinking is the wrong way round. Work done is a quantity that can be positive (increase KE) or negative (decrease KE). The mathematical representation you choose must reflect that. I don't understand the idea of choosing a model first (positive numbers only) and then demaning that the physical quantity can only be positive.

It's like demanding that the money in your bank account is always positive - and, then being confused about how to represent being overdrawn.

Choose the maths to fit the physics; not the other way round.

 

[Drakkith]

If the work done by a force is the consequence of the dot product of the force and the displacement of the particle, then it should be a scalar quantity which is true as well.

Certainly.

But then why does question of the work done being positive or negative (depending on the angle between the force vector and the displacement vector of the particle) even arise. Shouldn't we just be concerned only about it's magnitude like we do in the case of other scalar quantities like speed or K.E or distance??

No. Take the dot product of the vectors (10,0,0) and (-1,0,0).
The dot product is equal to -10. If these vectors represent the motion of an object and the applied force, then the force has done negative work on the object. This could slow the object down, or maintain the current velocity against another accelerating force (a rocket engine firing while a rocket falls towards a planet or moon under gravity for example), etc.

 

[PhDeezNutz]

1) how can work be negative?

Due to the work energy theorem the work done is equal to the change in kinetic energy. So if a box is moving across a surface friction will definitely slow it down and make it lose energy. Hence the work done by friction is negative. Cosine of 180 is -1.

2) Why does the angle matter when doing negative work?

Suppose you have a horizontal frictionless surface and you are somehow anchored to it. Further suppose a short box (shorter than you) is sliding past you at a speed $v$ and you throw a Batman grappling hook and it attaches to the box at an angle and the box slows down across the frictionless surface as you pull it. It is only the tension component that is horizontal will contribute to slowing down. This is why you need the dot product.

 

[haruspex]

Shouldn't we just be concerned only about its magnitude like we do in the case of other scalar quantities like speed or K.E or distance??

Don’t confuse scalars with magnitudes. Scalars can be negative, magnitudes can't.
Scalars are just like one dimensional vectors (but to a mathematician, technically they are not vectors).
Examples of scalars include changes to magnitudes. In °A, temperature is a magnitude, but the change to a temperature is a scalar. Likewise a change in energy.

 

[Manish_529]

Don’t confuse scalars with magnitudes. Scalars can be negative, magnitudes can't.
Scalars are just like one dimensional vectors (but to a mathematician, technically they are not vectors).
Examples of scalars include changes to magnitudes. In °A, temperature is a magnitude, but the change to a temperature is a scalar. Likewise a change in energy.

but in case of speed or distance we don't bother assigning positive or negative signs to them why is that?

 

[kuruman]

but in case of speed or distance we don't bother assigning positive or negative signs to them why is that?

Because speed and distance are respectively the magnitudes of vectors velocity and displacement. Vector magnitudes are by definition positive.

 

[PeroK]

but in case of speed or distance we don't bother assigning positive or negative signs to them why is that?

What's your definition of speed?

 

[kuruman]

Why does the angle matter when doing negative work?

The angle always matters and determines the sign of the work done by a force on an object. The expression for work $W$ done by a constant force $\mathbf F$ when the object is displaced by $\mathbf s$ is $$W=\mathbf F\cdot\mathbf s=Fs\cos\!\theta$$where
$F=~$ the magnitude of the force, always a positive number;
$s=~$ the magnitude of the displacement, always a positive number;
$\cos\!\theta=~$ the cosine of the angle between the force vector and the displacement vector.

Note that angle $\theta$ can be anything from 0° to 360° which means that the cosine of the angle can have any value between -1 and +1. Specifically, if $~0^{\circ} \leq \theta \leq 180^{\circ},$ the cosine is positive and if $~180^{\circ} \leq \theta \leq 360^{\circ},$ the cosine is negative.

 

[PhDeezNutz]

The angle always matters and determines the sign of the work done by a force on an object. The expression for work $W$ done by a constant force $\mathbf F$ when the object is displaced by $\mathbf s$ is $$W=\mathbf F\cdot\mathbf s=Fs\cos\!\theta$$where
$F=~$ the magnitude of the force, always a positive number;
$s=~$ the magnitude of the displacement, always a positive number;
$\cos\!\theta=~$ the cosine of the angle between the force vector and the displacement vector.

Note that angle $\theta$ can be anything from 0° to 360° which means that the cosine of the angle can have any value between -1 and +1. Specifically, if $~0^{\circ} \leq \theta \leq 180^{\circ},$ the cosine is positive and if $~180^{\circ} \leq \theta \leq 360^{\circ},$ the cosine is negative.

True. But I thought OP was specifically asking about negative work. Might have misread.

And I was more trying to contrive a situation in which negative work is done and the angle between the force and displacement wasn’t 180.

But of course you are correct.

 

[kuruman]

True. But I thought OP was specifically asking about negative work. Might have misread.

I wouldn't know whether you misread but OP correctly states

W=F.S (work done by a force is the dot product of the force and the displacement of the particle)

OP's question seems to stem from the misconception that @haruspex pointed out in post #5, that the dot product must be positive. The point of my post #9 is that the dot product is the algebraic product of three things, two which are always positive and one which could be positive or negative and cannot be ignored if one wishes to form the dot product correctly.

 

[Drakkith]

but in case of speed or distance we don't bother assigning positive or negative signs to them why is that?

Always, always, always look at the definition of terms like speed, distance, velocity, etc. Doing this any time you have a question about a concept will cut down your confusion immensely in my experience. Usually questions like, "Why is X the way it is" can immediately be answered with, "Because it is defined that way."

 

[haruspex]

haruspex pointed out in post #5, that the dot product must be positive

It was @Drakkith in post #3.

 

[kuruman]

It was @Drakkith in post #3.

I was referring to OP's misconception that prompted you to write

Don’t confuse scalars with magnitudes. Scalars can be negative, magnitudes can't.

which, in my opinion, is at the root of OP's misunderstanding.

 

[Manish_529]

But what does a negative work represent then? Cause using it in case of vectors like velocity it merely tells us about it's direction but, that won't make sense in case of work because it is an scalar quantity and doesn't need a sense of direction to defined

 

[PeroK]

But what does a negative work represent then? Cause using it in case of vectors like velocity it merely tells us about it's direction but, that won't make sense in case of work because it is an scalar quantity and doesn't need a sense of direction to defined

Plus and minus don't necessarily define a sense of direction. If you are a soccer fan, take goal difference, for example. You need to allow positive and negative goal differences, but goal difference isn't a vector.

Whereas, goals scored can't be negative.

 

[PhDeezNutz]

You’re asking some tough questions about scalars vs vectors. The definition usually given in introductory textbooks

I.e. vectors have direction and magnitude while scalars only have magnitude

Is sort of superficial because as you say “if something is negative then how can it be a magnitude?”

The more advanced (and correct) definition is that

“Scalars are unaffected by coordinate rotations”

“Vectors are affected by coordinate rotations”

additionally

“Both vectors and scalars can be affected by translating the coordinate system”

This is usually beyond the scope of an introductory physics course but good question.

 

[PeroK]

Plus and minus don't necessarily define a sense of direction. If you are a soccer fan, take goal difference, for example. You need to allow positive and negative goal differences, but goal difference isn't a vector.

Whereas, goals scored can't be negative.

If you want a better "physics" example, then mass (which can only be positive) is a scalar; and, electric charge (which can be positive or negative) is also a scalar.

But, an electric field or an electric force is a vector. And a vector has three components. The only confusion is that in a one-dimensional system, two of the directions can be suppressed. In which case the vector may be reduced to looking like a scalar, but it's not. It's still a vector and still has three components, even if two of those components are zero.

 

[Steve4Physics]

But what does a negative work represent then?

Maybe there is a basic concept about work that is causing you difficulty. This may help...

An object is moving to the right on a frictionless level surface.

You push the object to the right; that makes it speed up. Work done on the object is positive, say 100J. The change in the object’s kinetic energy is 100J (an increase)

You push the object to the left; that makes it slow down. Work done on the object is negative, say -100J. The change in the object’s kinetic energy is -100J (a decrease).

Positive work done (by a force on an object) means the force is delivering energy to the object.
Negative work done (by a force on an object) means the force is removing energy from the object.

 

[kuruman]

But what does a negative work represent then? Cause using it in case of vectors like velocity it merely tells us about it's direction but, that won't make sense in case of work because it is an scalar quantity and doesn't need a sense of direction to defined

By doing work on an object, a force changes the kinetic energy of the object.

Consider a single force acting on an object that is already moving. By Newton's first law the object must change its velocity.

• If the force is acting in the same direction as the velocity (a) the force does positive work on the object and (b) the speed and hence the kinetic energy of the object increases.
• If the force is acting in the opposite direction as the velocity (a) the force does negative work on the object and (b) the speed and hence the kinetic energy of the object decreases.
• If the force is acting in a perpendicular direction to the velocity (a) the force does zero work on the object and (b) the speed and hence the kinetic energy of the object stays the same.

To answer your question directly, negative work represents lost kinetic energy. By doing work on an object, a force mediates changes in the kinetic energy of an object. This change is bidirectional, positive or negative.

Real life example
You are in a car moving at 10 mph, forward or in reverse - it doesn't matter.

• If you push on the accelerator pedal, the engine provides a force in the direction of the velocity and the speed increases. The force exerted by the engine on the car does positive work that equals the extra joules that are added to the car's kinetic energy.
• If you push on the brake pedal, the engine provides a force opposite to the direction of the velocity and the speed decreases. The force exerted by the brakes on the car does negative work that equals the joules that are subtracted from the car's kinetic energy.
• If you only turn the wheel without pushing on any pedals, the force of friction at the tires provides a force perpendicular to the direction of the velocity and the speed does not change. The force exerted by friction on the car does zero work and no joules are subtracted from the car's kinetic energy.

Try it and see for yourself.