The force (on a single point particle) is also simply a function: $$\begin{cases} \mathbf F: \mathbb R \rightarrow \mathbb R^3\\ t \longmapsto \mathbf F(t) \end{cases}$$ In other words, $\mathbf F$ takes time as an input and spits out the force exerted on the particle at that time as its output.

However, I assume your confusion comes from the definition of work, usually stated as: $$W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$$ where $C_{AB}$ is the path that the particle takes while going from point $A$ to point $B$. However, with a bit of nitpicking, this notation may be somewhat misleading. I would choose to write the definition of work in this way instead: $$W_{A\rightarrow B} \equiv \int_{t_A}^{t_B}d\mathbf r(t) \cdot \mathbf F(t) \equiv \int_{t_A}^{t_B}dt \ \mathbf v(t) \cdot \mathbf F(t) \qquad (*)$$ where I used $d\mathbf r(t) = \frac{d\mathbf r}{dt} \ dt = \mathbf v(t) dt$. If you're wondering why people sometimes write $\mathbf F(\mathbf r)$, see the last part of my answer.

This implies that as a mathematical object, work can be thought of as a functional, which takes the two functions $\mathbf v: \mathbb R \rightarrow \mathbb R^3$ and $\mathbf F: \mathbb R \rightarrow \mathbb R^3$, and returns a real number. So if I call the set of "sufficiently well-behaved" (sorry this is as rigorous as I can get ;)) functions from $\mathbb R$ to $\mathbb R^3$ simply $\mathfrak F$, the work functional can be characterized as: $$\begin{cases} W: \mathfrak F \times \mathfrak F \rightarrow \mathbb R\\ (\mathbf v, \mathbf F) \longmapsto W[\mathbf v,\mathbf F] \end{cases}$$ In other words, the work functional takes the velocity function, or equivalently the position (disregarding a constant shift), as well as the force function, and spits out a number.

You might be asking why the force is usually written as $\mathbf F(\mathbf r)$, instead of $\mathbf F(t)$.This is because most of the time, people are thinking of a force field, which is technically a different mathematical object: $$\begin{cases} \mathbf f: \mathbb R^3 \rightarrow \mathbb R^3\\ \mathbf r \longmapsto \mathbf f(\mathbf r) \end{cases}$$ A force field takes a position vector $\mathbf r$, and spits out the force $\mathbf f$ exerted on a particle (with a specific mass, etc.) if it was sitting at $\mathbf r$. For example, a gravitational force field $\mathbf f(\mathbf r) = GMm \ {\mathbf r}/{|\mathbf r|^3}$ tells you that if you had a particle of mass $m$ at position $\mathbf r$, the force on that particle would be $\mathbf f(\mathbf r)$.

Now the force $\mathbf F: \mathbb R \rightarrow \mathbb R^3$ in the case of a force field is simply a composition of the two functions $\mathbf f: \mathbb R^3 \rightarrow \mathbb R^3$ and $\mathbf r: \mathbb R \rightarrow \mathbb R^3$. In other words, the force at time $t$ is simply the force field evaluated at the position at which the particle is at that instant. So: $$\mathbf F(t) \equiv \mathbf f(\mathbf r(t))$$ Using this on the definition of work $(*)$ you get: $$W_{A \rightarrow B} = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf f(\mathbf r(t))$$ which is what people actually mean by $W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$.

Keep in mind that these are mostly just mathematical (as well as notational) nitpicking. They don't really affect any of the actual Physics in the problem :).

The force (on a single point particle) is also simply a function: $$\begin{cases} \mathbf F: \mathbb R \rightarrow \mathbb R^3\\ t \longmapsto \mathbf F(t) \end{cases}$$ In other words, $\mathbf F$ takes time as an input and spits out the force exerted on the particle at that time as its output.

However, I assume your confusion comes from the definition of work, usually stated as: $$W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$$ where $C_{AB}$ is the path that the particle takes while going from point $A$ to point $B$. However, with a bit of nitpicking, this notation may be somewhat misleading. I would choose to write the definition of work in this way instead: $$W_{A\rightarrow B} \equiv \int_{t_A}^{t_B}d\mathbf r(t) \cdot \mathbf F(t) \equiv \int_{t_A}^{t_B}dt \ \mathbf v(t) \cdot \mathbf F(t) \qquad (*)$$ where I used $d\mathbf r(t) = \frac{d\mathbf r}{dt} \ dt = \mathbf v(t) dt$. If you're wondering why people sometimes write $\mathbf F(\mathbf r)$, see the last part of my answer.

This implies that as a mathematical object, work can be thought of as a functional, which takes the two functions $\mathbf v: \mathbb R \rightarrow \mathbb R^3$ and $\mathbf F: \mathbb R \rightarrow \mathbb R^3$, and returns a real number. So if I call the set of "sufficiently well-behaved" (sorry this is as rigorous as I can get ;)) functions from $\mathbb R$ to $\mathbb R^3$ simply $\mathfrak F$, the work functional can be characterized as: $$\begin{cases} W: \mathfrak F \times \mathfrak F \rightarrow \mathbb R\\ (\mathbf v, \mathbf F) \longmapsto W[\mathbf v,\mathbf F] \end{cases}$$ In other words, the work functional takes the velocity function, or equivalently the position (disregarding a constant shift), as well as the force function, and spits out a number.

You might be asking why the force is usually written as $\mathbf F(\mathbf r)$, instead of $\mathbf F(t)$.This is because most of the time, people are thinking of a force field, which is technically a different mathematical object: $$\begin{cases} \mathbf f: \mathbb R^3 \rightarrow \mathbb R^3\\ \mathbf r \longmapsto \mathbf f(\mathbf r) \end{cases}$$ A force field takes a position vector $\mathbf r$, and spits out the force $\mathbf f$ exerted on a particle (with a specific mass, etc.) if it was sitting at $\mathbf r$. For example, a gravitational force field $\mathbf f(\mathbf r) = GMm \ {\mathbf r}/{|\mathbf r|^3}$ tells you that if you had a particle of mass $m$ at position $\mathbf r$, the force on that particle would be $\mathbf f(\mathbf r)$.

Now the force $\mathbf F: \mathbb R \rightarrow \mathbb R^3$ in the case of a force field is simply a composition of the two functions $\mathbf f: \mathbb R^3 \rightarrow \mathbb R^3$ and $\mathbf r: \mathbb R \rightarrow \mathbb R^3$. In other words, the force at time $t$ is simply the force field evaluated at the position at which the particle is at that instant. So: $$\mathbf F(t) \equiv \mathbf f(\mathbf r(t))$$ Using this on the definition of work $(*)$ you get: $$W_{A \rightarrow B} = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf f(\mathbf r(t))$$ which is what people actually mean by $W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$.

Keep in mind that these are mostly just mathematical (as well as notational) nitpicking. They don't really affect any of the actual Physics in the problem :).

Response to comment:

Just as I talked about a position-dependent force field $\mathbf f(\mathbf r)$, you can have a more generalized time,velocity and position-dependent force field $\mathbf f(\mathbf r,\mathbf v,t)$.

For example, consider a 1D spring with velocity-dependent friction, having a time dependent spring constant $k(t)$. The force exerted on a particle by this spring is simply $-k(t) x-\gamma v$. So you can define a generalized force field $f$ as $f(x,v,t) \equiv -k(t) x-\gamma v$ ($\gamma$ is the friction constant).

More generally you can have a generalized force field $\mathbf f$: $$\begin{cases} \mathbf f: \mathbb R^3\times\mathbb R^3 \times \mathbb R \ \rightarrow \mathbb R^3\\ (\mathbf r, \mathbf v, t) \longmapsto \mathbf f(\mathbf r,\mathbf v,t) \end{cases}$$ Again, the force experienced by a particle in such a force field would be a composition of $\mathbf f$ with the functions $\mathbf r$ and $ \mathbf v$; i.e. $$\mathbf F(t) \equiv \mathbf f(\mathbf r(t),\mathbf v(t),t)$$ So the force experienced by the particle is both explicitly dependent on time, as well as varying through the change of the particles position and velocity.

The work for this force field on the particle would simply be: $$W_{A \rightarrow B} = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf F(t) = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf f(\mathbf r(t),\mathbf v(t),t)$$

Note that in principle, you can even think of more and more complicated force fields, the main idea is however the same.

The force (on a single point particle) is also simply a function: $$\begin{cases} \mathbf F: \mathbb R \rightarrow \mathbb R^3\\ t \longmapsto \mathbf F(t) \end{cases}$$ In other words, $\mathbf F$ takes time as an input and spits out the force exerted on the particle at that time as its output.

However, I assume your confusion comes from the definition of work, usually stated as: $$W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$$ where $C_{AB}$ is the path that the particle takes while going from point $A$ to point $B$. However, with a bit of nitpicking, this notation may be somewhat misleading. I would choose to write the definition of work in this way instead: $$W_{A\rightarrow B} \equiv \int_{t_A}^{t_B}d\mathbf r(t) \cdot \mathbf F(t) \equiv \int_{t_A}^{t_B}dt \ \mathbf v(t) \cdot \mathbf F(t) \qquad (*)$$ where I used $d\mathbf r(t) = \frac{d\mathbf r}{dt} \ dt = \mathbf v(t) dt$. If you're wondering why people sometimes write $\mathbf F(\mathbf r)$, see the last part of my answer.

This implies that as a mathematical object, work can be thought of as a functional, which takes the two functions $\mathbf v: \mathbb R \rightarrow \mathbb R^3$ and $\mathbf F: \mathbb R \rightarrow \mathbb R^3$, and returns a real number. So if I call the set of "sufficiently well-behaved" (sorry this is as rigorous as I can get ;)) functions from $\mathbb R$ to $\mathbb R^3$ simply $\mathfrak F$, the work functional can be characterized as: $$\begin{cases} W: \mathfrak F \times \mathfrak F \rightarrow \mathbb R\\ (\mathbf v, \mathbf F) \longmapsto W[\mathbf v,\mathbf F] \end{cases}$$ In other words, the work functional takes the velocity function, or equivalently the position (disregarding a constant shift), as well as the force function, and spits out a number.

You might be asking why the force is usually written as $\mathbf F(\mathbf r)$, instead of $\mathbf F(t)$.This is because most of the time, people are thinking of a force field, which is technically a different mathematical object: $$\begin{cases} \mathbf f: \mathbb R^3 \rightarrow \mathbb R^3\\ \mathbf r \longmapsto \mathbf f(\mathbf r) \end{cases}$$ A force field takes a position vector $\mathbf r$, and spits out the force $\mathbf f$ exerted on a particle (with a specific mass, etc.) if it was sitting at $\mathbf r$. For example, a gravitational force field $\mathbf f(\mathbf r) = GMm \ {\mathbf r}/{|\mathbf r|^3}$ tells you that if you had a particle of mass $m$ at position $\mathbf r$, the force on that particle would be $\mathbf f(\mathbf r)$.

Now the force $\mathbf F: \mathbb R \rightarrow \mathbb R^3$ in the case of a force field is simply a composition of the two functions $\mathbf f: \mathbb R^3 \rightarrow \mathbb R^3$ and $\mathbf r: \mathbb R \rightarrow \mathbb R^3$. In other words, the force at time $t$ is simply the force field evaluated at the position at which the particle is at that instant. So: $$\mathbf F(t) \equiv \mathbf f(\mathbf r(t))$$ Using this on the definition of work $(*)$ you get: $$W_{A \rightarrow B} = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf f(\mathbf r(t))$$ which is what people actually mean by $W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$.

The force (on a single point particle) is also simply a function: $$\begin{cases} \mathbf F: \mathbb R \rightarrow \mathbb R^3\\ t \longmapsto \mathbf F(t) \end{cases}$$ In other words, $\mathbf F$ takes time as an input and spits out the force exerted on the particle at that time as its output.

However, I assume your confusion comes from the definition of work, usually stated as: $$W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$$ where $C_{AB}$ is the path that the particle takes while going from point $A$ to point $B$. However, with a bit of nitpicking, this notation may be somewhat misleading. I would choose to write the definition of work in this way instead: $$W_{A\rightarrow B} \equiv \int_{t_A}^{t_B}d\mathbf r(t) \cdot \mathbf F(t) \equiv \int_{t_A}^{t_B}dt \ \mathbf v(t) \cdot \mathbf F(t) \qquad (*)$$ where I used $d\mathbf r(t) = \frac{d\mathbf r}{dt} \ dt = \mathbf v(t) dt$. If you're wondering why people sometimes write $\mathbf F(\mathbf r)$, see the last part of my answer.

This implies that as a mathematical object, work can be thought of as a functional, which takes the two functions $\mathbf v: \mathbb R \rightarrow \mathbb R^3$ and $\mathbf F: \mathbb R \rightarrow \mathbb R^3$, and returns a real number. So if I call the set of "sufficiently well-behaved" (sorry this is as rigorous as I can get ;)) functions from $\mathbb R$ to $\mathbb R^3$ simply $\mathfrak F$, the work functional can be characterized as: $$\begin{cases} W: \mathfrak F \times \mathfrak F \rightarrow \mathbb R\\ (\mathbf v, \mathbf F) \longmapsto W[\mathbf v,\mathbf F] \end{cases}$$ In other words, the work functional takes the velocity function, or equivalently the position (disregarding a constant shift), as well as the force function, and spits out a number.

You might be asking why the force is usually written as $\mathbf F(\mathbf r)$, instead of $\mathbf F(t)$.This is because most of the time, people are thinking of a force field, which is technically a different mathematical object: $$\begin{cases} \mathbf f: \mathbb R^3 \rightarrow \mathbb R^3\\ \mathbf r \longmapsto \mathbf f(\mathbf r) \end{cases}$$ A force field takes a position vector $\mathbf r$, and spits out the force $\mathbf f$ exerted on a particle (with a specific mass, etc.) if it was sitting at $\mathbf r$. For example, a gravitational force field $\mathbf f(\mathbf r) = GMm \ {\mathbf r}/{|\mathbf r|^3}$ tells you that if you had a particle of mass $m$ at position $\mathbf r$, the force on that particle would be $\mathbf f(\mathbf r)$.

Now the force $\mathbf F: \mathbb R \rightarrow \mathbb R^3$ in the case of a force field is simply a composition of the two functions $\mathbf f: \mathbb R^3 \rightarrow \mathbb R^3$ and $\mathbf r: \mathbb R \rightarrow \mathbb R^3$. In other words, the force at time $t$ is simply the force field evaluated at the position at which the particle is at that instant. So: $$\mathbf F(t) \equiv \mathbf f(\mathbf r(t))$$ Using this on the definition of work $(*)$ you get: $$W_{A \rightarrow B} = \int_{t_A}^{t_B} d\mathbf r(t) \cdot \mathbf f(\mathbf r(t))$$ which is what people actually mean by $W_{A\rightarrow B} \equiv \int_{C_{AB}}d\mathbf r \cdot \mathbf F(\mathbf r)$.

Keep in mind that these are mostly just mathematical (as well as notational) nitpicking. They don't really affect any of the actual Physics in the problem :).