How can there be force with no work if $W = Fd$ and $F = ma$? In order to measure a force we need to know the mass of the object the force is applied to and the acceleration of said object in the direction of the force, right? So, if there is a force, that means there is acceleration of an object, if there is acceleration there is movement, if there is movement there is distance, so, therefore, there is work since $W = Fd$. So, how can you have a force with no work?
 A: 
The acceleration of said object in the direction of the force, right?

Yes!

So, if there is a force, that means there is acceleration of an object

If there is a net force, then yes!

if there is acceleration there is movement

Usually, yes, although you can have a non-zero instantaneous acceleration with a zero instantaneous velocity (e.g. a ball thrown straight up at the apex of its trajectory), but at the next instant the velocity will be non-zero. So basically yes, non-zero acceleration over a finite time will give motion.

if there is movement there is distance

Yes! If velocity is non-zero then the change in position is non-zero

so, therefore, there is work since $W=Fd$.

And here is the pitfall so many students who are new to physics fall into. In general, work is not $W=Fd$. This is only true when the force has a constant magnitude $F$ and points along the displacement along the entire path in question (whose path length is $d$). If these conditions are not met you have to go to the more general definition of the work done by a force $\mathbf F$ along some path $C$
$$W=\int_C\mathbf F\cdot\text d\mathbf x$$
Note that this integral involves adding up all of the infinitesimal dot products $\mathbf F\cdot\text d\mathbf x$, where $\text d\mathbf x$ points along the path. This leads to the main question:

So, how can you have a force with no work?

Just have the force always be perpendicular to the path. Then the work done is $0$, since $\mathbf F\cdot\text d\mathbf x=0$. An easy example of this is uniform circular motion, like a ball on a string moving in a circle at a constant speed. The tension in the string does no work, even though the tension is responsible for the motion of the ball. Another example is the normal force acting on a block sliding down an incline; the normal force is perpendicular to the displacement of the block, and hence it does no work.
A: 
So, if there is a force, that means there is acceleration of an object

This is not always true. The $F$ in $F=ma$ is the net force.

how can you have a force with no work?

One common case is when an equal and opposite force opposes it. For example, gravity does no work on an object on a table. Neither do you do (mechanical) work by pushing against a wall.
A: In physics it is very important to not just have a formula but to know exactly what the variables in the formula mean. The meaning is very precise.
In Newton’s 2nd law we have $\Sigma \vec F = m \vec a$. In this formula, $\Sigma \vec F$ is the net external force acting on the system. $m$ is the mass of the system. And $\vec a$ is the acceleration of the center of mass of the system.
In the definition of work we have $W=\vec F \cdot \vec d$. In this formula, $\vec F$ is a single external force acting on the system. $W$ is the mechanical work done on the system by that force. And $\vec d$ is the displacement of the material of the system at the point of application of the force.
These definitions are very specific and many problems arise if you deviate in their usage.

So, if there is a force, that means there is acceleration of an object, if there is acceleration there is movement, if there is movement there is distance

If there is a force, then there is acceleration of the center of mass of the system. If there is acceleration of the center of mass then there is movement of the center of mass of the system.
However, if an object is non-rigid then the motion of the center of mass of the system need not coincide with the displacement of some part of the system. Specifically, it is common for a force to accelerate the center of mass of a system while the material at the point of application of the force is stationary. This means that the force does no work on the system.
Examples include cars, jumping, and collisions with stationary objects. In all cases the force producing the acceleration does no work and instead there is an conversion of internal energy that provides any necessary energy.
A: Don't forget the summation symbol in Newton's second law. It is not $F=ma$ but rather:
$$\sum F=ma$$
If you ever see just $F=ma$, then the summation is implied and the $F$ is not a single force but rather the sum of all forces.
So, the $F$ in $F=ma$ is not the same as the $F$ in $W=Fd$ and thus there is no conflict. In words: It is not a single force that causes acceleration, only the sum of all forces. So a single force does not necessarily cause acceleration and displacement and thus doesn't necessarily do work.
A: If there is a net force, there will be a net acceleration.
Acceleration means there is going to be displacement, and now there is a possibility of work.
Possibility because though force dictates current acceleration as per $\vec{F} = m\vec{a}$, it doesn't have control over current displacement.
Current displacement is a result of current velocity and if the force is hell bent upon keeping its direction perpendicular to the current velocity, the displacement will always be perpendicular to the force (for the duration of the force).
And in that case, there will be no work done.
Another way to look at it is this. While a net force will always cause a change in momentum, it may or may not cause a change in energy of the body on which it is applied.
The net change in momentum helps us determine the force using the simplified formula $\vec{F} = m\vec{a}$.
But, if this net change in momentum translates to only a change in direction of velocity and not its magnitude, there will be no change in energy of the body, and hence no work done.
