Positive vs. negative work in an elevator?

In summary: Likewise, for the x-component: \vec{F} = F_x d_x + F_y d_y + F_z d_z So, the x-component of the dot product is just the sum of the x-components of the individual forces. Now, if we were to evaluate the dot product in terms of magnitude, we'd get: \vec{F} \cdot \vec{d} = |F| \hat{j} Which is just the magnitude of the force |F|.
  • #1

Homework Statement



You are moving into an apartment and take the elevator to the 6th floor. Does the force exerted on you by the elevator do positive or negative work when the elevator (a) goes up and (b) goes down? Explain your answers.

Suppose your weight is 685N and that of your belongings is 915N. (a) Determine the work done by the elevator in lifting you and your belongings up to the 6th floor (15.2m) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity?


Homework Equations



W=Fdcos(theta)

The Attempt at a Solution



For the concept question, I know that work is negative when it is going in the opposite direction of the force, but I'm thrown off by which force. The question indicates that there is no acceleration, so there is a gravitational pull downwards equal to a normal force upwards. So which force is supposed to be opposed?

As for the second part, this is what I got:
685+915=1600N
1600(15.2)(cos0)=24320J
685(15.2)(cos180)=-10412J
BUT here's what confuses me - since displacement is a vector quantity, shouldn't one of the above equations have a negative displacement since it's going in the opposite direction?

Argh, I've been mulling over this and I'm still so confused. Please help!
 
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  • #2
Welcome to PF thepegmeister,

thepegmeister said:
For the concept question, I know that work is negative when it is going in the opposite direction of the force, but I'm thrown off by which force.

Here is the answer to that question:


thepegmeister said:

Homework Statement



You are moving into an apartment and take the elevator to the 6th floor. Does the force exerted on you by the elevator do positive or negative work when the elevator (a) goes up and (b) goes down? Explain your answers.

Say you ascend at a constant speed. You're right that 0 total work is done on you for most of the ride, because the work done depends on the net force on you, and if the normal force on you from the elevator floor just balances your weight, then the net force on you is 0.

However, the work done by the elevator normal force is non-zero and positive, since it exerts a force on you that is in the same direction as your displacement. It's just that this positive work done by the elevator on you happens to be exactly canceled by the negative work done on you by gravity.

Bear in mind also, that in order to even get you moving in the first place, the elevator initially has to exert an upward normal force on you that is higher than your weight, so that there will be a net upward force on you, and hence an upward acceleration. It is only once you have accelerated up to a certain speed that the elevator force is lessened until it just balances your weight and you continue to ascend at a constant speed (which is much more comfortable).

So, in other words, at the very beginning of the ride, a non-zero and positive amount of work is done on you by the elevator in order to get you in motion in the first place.

That's the conceptual summary for the ascent. Now can you do it for the descent?
 
  • #3
thepegmeister said:
As for the second part, this is what I got:
685+915=1600N
1600(15.2)(cos0)=24320J
685(15.2)(cos180)=-10412J
BUT here's what confuses me - since displacement is a vector quantity, shouldn't one of the above equations have a negative displacement since it's going in the opposite direction?

The full equation for work is [itex] W = \vec{F} \cdot \vec{d} [/itex]. In other words, the work is the dot product of the force vector with the displacement vector. There are TWO ways to evaluate the dot product. The first way is component-wise. If you resolve each vector into x, y, and z components, you get:

[tex] \vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} [/tex]

[tex] \vec{d} = d_x \hat{i} + d_y \hat{j} + d_z \hat{k} [/tex]

where i, j, and k are unit vectors in the x, y, and z directions respectively. The components dx, dy, and dz are the displacements in the x, y, and z directions respectively, and they can each be either positive or negative. Same with the individual components of the force So, anyway, using components, the definition of the dot product is:

[tex] \vec{F} \cdot \vec{d} = F_x d_x + F_y d_y + F_z d_z [/tex]Now, it turns out, that this is a one-dimensional problem. Both vectors are in the "vertical" direction, which we'll call the y-direction. Therefore, the vectors reduce to:

[tex] \vec{F} = F \hat{j} [/tex]

[tex] \vec{d} = -d \hat{j} [/tex]

Where [itex] F = |\vec{F}| [/itex] and [itex] d = |\vec{d}| [/itex]. In other words, F and d are the magnitudes of their respective vectors. In other words, we're saying that the y-component of the force was the entire force, and that it was in the positive y-direction: [itex] F_y = |\vec{F}| [/itex]. We're also saying that the y-component of the displacement was, in fact, the entire displacement, and it happened to be in the negative y-direction: [itex] d_y = -|\vec{d}| [/itex]. With that being done, the dot product becomes:

[tex] \vec{F} \cdot \vec{d} = F_y d_y = F(-d) = -Fd [/tex]

With this first method, we explicitly included the negative sign on the displacement vector to indicate that it was in the negative y-direction.

The second way to evaluate the dot product (which is the way you are doing it) is to use the fact that the dot product is ALSO given by the expression:

[tex] \vec{F} \cdot \vec{d} = |\vec{F}||\vec{d}|\cos\theta = Fd\cos\theta [/tex]

where theta is the angle between the two vectors. I put in the middle part in the equation above to emphasize that, by definition, F and d here are the magnitudes of the vectors. As a result, they are both positive, and explicit signs indicating direction are not put in. Using this method, the fact that the vectors are in opposite directions is already taken care of by the cosine factor, since cos(180°) = -1.

That was really long-winded, and the actual answer to your question was just the statement in bold and italics above.
 
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  • #4
Thank you guys so much! I wish they would explain these things in the textbook - the small details always get me.
 
  • #5




The force exerted on you by the elevator does positive work when the elevator goes up and negative work when the elevator goes down. This is because in both cases, the force exerted by the elevator is in the same direction as your displacement when going up, and in the opposite direction when going down. Therefore, the force and displacement are in the same direction when the elevator goes up, resulting in positive work, and in opposite directions when the elevator goes down, resulting in negative work.

For the calculations, you are correct in using the formula W=Fdcos(theta). However, the force and displacement should have opposite signs when going up and down. This is because when the elevator goes up, the force and displacement are in the same direction, so the angle between them is 0 degrees, resulting in cos(0) = 1. When the elevator goes down, the force and displacement are in opposite directions, so the angle between them is 180 degrees, resulting in cos(180) = -1. Therefore, the correct calculations should be:

a) W = (685N + 915N)(15.2m)(cos0) = 24320J
b) W = 685N(15.2m)(cos180) = -10412J

In both cases, the work done by the elevator is positive, since the force and displacement are in the same direction when going up, and in opposite directions when going down. The negative sign in the second calculation indicates that the work is done in the opposite direction of the displacement, which is downwards.
 

1. What is the difference between positive and negative work in an elevator?

Positive work in an elevator refers to the work done by the elevator on an object as it moves upwards. This requires force to be applied in the same direction as the displacement. On the other hand, negative work in an elevator occurs when the elevator moves downwards and the force applied is in the opposite direction of the displacement.

2. How is positive work calculated in an elevator?

Positive work is calculated by multiplying the force applied on an object by the distance it travels in the direction of the force. In an elevator, this would be the force needed to lift the object and the distance it moves upwards.

3. Can negative work be done in an elevator?

Yes, negative work can be done in an elevator when the elevator is moving downwards and the force applied is in the opposite direction of the displacement. This can happen when the elevator is descending with an object that is heavier than the force applied.

4. How does work affect an object in an elevator?

The work done on an object in an elevator can change its kinetic energy, potential energy, or both. Positive work increases the object's kinetic or potential energy, while negative work decreases it.

5. Is positive work always better than negative work in an elevator?

It depends on the situation. Positive work can be beneficial as it can increase an object's energy, but too much positive work can also cause the object to gain too much kinetic energy and potentially cause damage. Negative work, on the other hand, can be beneficial in slowing an object down, but too much negative work can also slow an object down too quickly and potentially cause harm.

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