# Confused about work-energy

• StephenPrivitera
In summary: If there are no other forces acting on the object, then W_net = delta K. If you're lifting the box at a constant rate then the net force is zero so that no work is being done on the object.

#### StephenPrivitera

In my text, we learn that the work done on an object is equal to the change in kinetic energy of that object. It seems to me that this statement requires some restrictions. If I take a an object at rest of mass m and lift to a height h and place it at rest on a shelf, the work I've done is -mgh. But I haven't changed the kinetic energy.
If you consider some U-K system (sorry), perhaps where U is given by mgh, then the work done by the conservative force which provides U is always equal to the negative of the change in kinetic energy, provided there are no other forces.
In the first example, the total energy changes and W=deltaK fails. In the second, the total energy is constant and W=deltaK holds. How do I know in general when to apply the work-energy theorem??

Your text isn't giving you the whole story.

The work done on an object = the sum of all other energies. That included Gravitational Potential energy (from your example), thermal energy, electrical energy, kinetic energy, ...

You're definitely on the right track since you saw there was something askew.

I might be misunderstanding you, enigma, but what you said sounds a bit off to my classical ear. The work energy theorem states that the total (net) work on an object is equal to the change in the kinetic energy.

When I lift an object to place it on a high shelf, I do +mgh of work on it, but at the same time gravity does -mgh. The net work is zero, so the change in KE is zero. If I were to apply a force greater than the object's weight, then after I lift it a certain height , h, it will be moving with a certain KE, and continue to rise to another height.

When friction is present, it also does negative work. So Wnet=Wapplied-Wgravity-Wfriction

Any other forces that are on the object at the same time will do work on the object according to the relationship W=Fd cos [theta].

Stephen, I am assuming that inclusion of thermodynamics and electrical effects are a tiny bit beyond your current location in the text? Ultimately these must be considered as well, as enigma mentions.

THis work-energy theorem is actually a restatement of the Law of Conservation of energy. Sometimes it's easier to examine problems by comparing initial and final energy states.

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Originally posted by Chi Meson
The work energy theorem states that the total (net) work on an object is equal to the change in the kinetic energy.
Yes! I always screw that part up. I always forget that it's the net work we're dealing with.
So if a motor elevates an elevator initially at rest 1/2a meters in one second with an acceleration of a meters per second squared, the net work W on the elevator is 1/2ma2 since the velocity changes from 0 to a. The work done by gravity is -mga/2. So the work done by the motor Wm is found by solving W=Wm+Wg for W_m. We find W_m=1/2ma(a+g).

Is it also true that W_net=-[del]U?
Well, -[del]U=mga/2 and W_net=1/2ma2
So g=a? I don't know about that...
What are the conditions regarding W=-[del]U?

Originally posted by StephenPrivitera
In my text, we learn that the work done on an object is equal to the change in kinetic energy of that object. It seems to me that this statement requires some restrictions. If I take a an object at rest of mass m and lift to a height h and place it at rest on a shelf, the work I've done is -mgh. But I haven't changed the kinetic energy.
If you consider some U-K system (sorry), perhaps where U is given by mgh, then the work done by the conservative force which provides U is always equal to the negative of the change in kinetic energy, provided there are no other forces.
In the first example, the total energy changes and W=deltaK fails. In the second, the total energy is constant and W=deltaK holds. How do I know in general when to apply the work-energy theorem??

If you're lifting the box at a constant rate then the net force is zero so that no work is being done on the object.

Pete

>>Is it also true that W_net=-U?
Well, -U=mga/2 and W_net=1/2ma2
So g=a? I don't know about that...
What are the conditions regarding W=-U?<<

No.

I really have a hard time with the subscripts in this format, so I'll use the style you're using. Please excuse in advance my style if it seems pedantic.

The work-energy theorem states W_net = delta K .

In translation, W_net is the sum of each of the works done by all forces on an object during a time interval. Let's say all the works from non-perpendicular forces are: from an applied force (some guy pushing), W_p ; friction, -W_f ; gravity, -W_g.

If a cart is pushed up a ramp, and it had an initial velocity, it would have an initial and a final KE, K_o and K_f. It would have an initial and final PE, U_o and U_f.

Compare the initial and final energy states:
K_o + U_o + W_p = K_f + U_f + Q

W_net = W_p - W_f - W_g
W_g = delta U = U_f - U_o
Friction turns kinetic energy into thermal energy, so -W_f = -Q
delta K = K_f - K_o

Substitute and manipulate, you get

W_net = delta K

## 1. What is work-energy?

Work-energy is a concept in physics that describes the relationship between the work done on an object and the resulting change in its kinetic energy. In simpler terms, it is the measure of the amount of energy needed to move an object from one position to another.

## 2. How is work-energy related to force?

Work-energy is directly related to force through the work-energy theorem, which states that the work done by a net force on an object is equal to the change in the object's kinetic energy. This means that the more force applied to an object, the more work is done and the greater the change in its kinetic energy.

## 3. What are some examples of work-energy in everyday life?

Some common examples of work-energy in everyday life include pushing a shopping cart, riding a bike, lifting weights, and throwing a ball. In each of these scenarios, work is being done to move an object from one position to another, resulting in a change in its kinetic energy.

## 4. How can work-energy be calculated?

The work done on an object can be calculated by multiplying the force applied to the object by the distance it moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance. The change in kinetic energy can then be calculated by using the formula KE = 1/2mv^2, where m is the mass of the object and v is its velocity.

## 5. What is the significance of work-energy in physics?

Work-energy is a fundamental concept in physics that helps us understand and analyze the motion of objects. It allows us to quantify the amount of energy needed to move an object and helps us predict how objects will behave in different situations. Work-energy is also closely related to other important concepts such as force, power, and momentum, making it an essential part of understanding the physical world.