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Relation to KE:

Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work

$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$

$$W= \frac{1}{2}m|\vec{v}|^2$$

For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$

This is only true for a constant force and an object moving in a straight line

Generally:

$$W= \int \vec{F} \cdot \vec{dr}$$

Situation:

Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$

Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$\frac{d\vec{r}(t)}{dt} = \frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k$

$d\vec{r}(t) = (\frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k) dt $

This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt

Vector field:

$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form

$X(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$

Dot product:

The qauntity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr

The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.

The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.

Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.

Integral:

Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.

ps:

When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form $ \vec{F}(\vec{r}(t))$ this is because we are evaluating the field at the position of the object at some time t

Relation to KE:

Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work

$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$

$$W= \frac{1}{2}m|\vec{v}|^2$$

For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$

This is only true for a constant force and an object moving in a straight line

Generally:

$$W= \int \vec{F} \cdot \vec{dr}$$

Situation:

Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$

Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$\frac{d\vec{r}(t)}{dt} = \frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k$

$d\vec{r}(t) = (\frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k) dt $

This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt

Vector field:

$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form

$x(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$

Dot product:

The quantity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr

The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.

The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.

Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.

Integral:

Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.

ps:

When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form $ \vec{F}(\vec{r}(t))$ this is because we are evaluating the field at the position of the object at some time t

Relation to KE:

Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work

$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$

$$W= \frac{1}{2}m|\vec{v}|^2$$

For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$

This is only true for a constant force and an object moving in a straight line

Generally:

$$W= \int \vec{F} \cdot \vec{dr}$$

Situation:

Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$

Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$\frac{d\vec{r}(t)}{dt} = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$d\vec{r}(t) = (x(t)\hat i +y(t)\hat j +z(t)\hat k)dt$

This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt

Vector field:

$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form

$X(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$

Dot product:

The qauntity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr

The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.

The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.

Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.

Integral:

Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.

ps:

When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form $ \vec{F}(\vec{r}(t))$ this is because we are evaluating the field at the position of the object at some time t

Relation to KE:

Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work

$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$

$$W= \frac{1}{2}m|\vec{v}|^2$$

For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$

This is only true for a constant force and an object moving in a straight line

Generally:

$$W= \int \vec{F} \cdot \vec{dr}$$

Situation:

Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$

Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$\frac{d\vec{r}(t)}{dt} = \frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k$

$d\vec{r}(t) = (\frac{dx(t)}{dt}\hat i +\frac{dy(t)}{dt}\hat j +\frac{dz(t)}{dt}\hat k) dt $

This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt

Vector field:

$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form

$X(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$

Dot product:

The qauntity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr

The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.

The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.

Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.

Integral:

Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.

ps:

When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form $ \vec{F}(\vec{r}(t))$ this is because we are evaluating the field at the position of the object at some time t

Relation to KE:

Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work

$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$

$$W= \frac{1}{2}m|\vec{v}|^2$$

For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$

This is only true for a constant force and an object moving in a straight line

Generally:

$$W= \int \vec{F} \cdot \vec{dr}$$

Situation:

Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$

Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$\frac{d\vec{r}(t)}{dt} = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$d\vec{r}(t) = (x(t)\hat i +y(t)\hat j +z(t)\hat k)dt$

This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt

Vector field:

$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form

$X(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$

Dot product:

The qauntity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr

The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.

The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.

Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.

Integral:

Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.

ps:

When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form$ \vec{F}(\vec{r}(t))$ this is because we are evaluation the field at the position of the object at some time t

Relation to KE:

Work being done on an object = the change in the objects kinetic energy. This result can be derived from the definition of work

$$W= \int \vec{F} \cdot \vec{dr}$$ $$W= \int \vec{F} \cdot \vec{v}dt$$ $$W= \int m\frac{d\vec{v}}{dt} \cdot \vec{v}dt$$

$$W= \frac{1}{2}m|\vec{v}|^2$$

For a constant force, for an object moving in a straight line, Work is Force×displacement. $W = |\vec{F}||\vec{x}|$

This is only true for a constant force and an object moving in a straight line

Generally:

$$W= \int \vec{F} \cdot \vec{dr}$$

Situation:

Given I have some object that I know follows some specific PREDEFINED path, $\vec{r}(t)$. As it moves it is in the presence of some vector field $\vec{F}$

Differential line element: $\vec{r}(t) = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$\frac{d\vec{r}(t)}{dt} = x(t)\hat i +y(t)\hat j +z(t)\hat k$

$d\vec{r}(t) = (x(t)\hat i +y(t)\hat j +z(t)\hat k)dt$

This represents an infinitesimal displacement that is traversed by the function $\vec{r}(t)$ in time dt

Vector field:

$\vec{F}$ is a vector field that has an assigned vector to every point in space usually in the form

$X(x,y,z)\hat i + y(x,y,z)\hat j + z(x,y,z)\hat k$

Dot product:

The qauntity $\vec{F} \cdot \vec{dr}$ represents the infinitesimal amount of work being done on an object moving a distance of dr

The dot product between $\vec{F}$ and $\vec{dr}(t)$ is a measure of the alignment between these 2 vectors. If at a point in space a force vector points in the direction of the movement of the object, more work is done than if the force points perpendicular to the direction of motion.

The way I like to think about line integrals is that you are DEFINING a specific path that the object is taking. If at a point in time, the force is pointing perpendicular to the motion of an object, then no work can be done. Because if work WAS being done, then that force would cause a component of the displacement to be in the direction of the force( because it would be accelerating). So because we are defining a specific path, we are saying that IF an object WERE to be going on this path, then how much work would a force be doing on an object.

Another way to get a feel for why there's a dot product, is that that if a ball ORIGINALLY is moving in a straight line, and I exert a force perpendicular to the motion of the object, the force would cause an acceleration, changing the path of the ball to be curved, this dot product would "pick out" the displacement caused by the force(as it find the component in the direction of force), instead of adding up the original velocity of the ball.

Integral:

Because F and dr are changing as the ball moves we use an integral to sum up the contributions of all the $\vec{F} \cdot \vec{dr}$ elements as the line integral iterates in time to find the total amount of work done on the object.

ps:

When computing the line integral make sure that $\vec{F}$ is a function of the path $\vec{r}(t)$ in the form  $ \vec{F}(\vec{r}(t))$ this is because we are evaluating the field at the position of the object at some time t