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?
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$\begingroup$ This might help. Is work done in rolling friction? $\endgroup$– mmesser314Commented Jul 8, 2021 at 4:22
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$\begingroup$ try reading this hyperphysics.phy-astr.gsu.edu/hbase/wcon.html $\endgroup$– Mr AndersonCommented Jul 9, 2021 at 6:55
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5 Answers
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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.
answered Jul 8, 2021 at 3:52
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1$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$– rob ♦Commented Jul 9, 2021 at 15:50
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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.
answered Jul 8, 2021 at 2:43
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$\begingroup$ Okay, so if a bullet hits a wall and becomes flattened, did the wall exert a force on the bullet? If it did, how do we measure that force since it didn't cause the bullet to accelerate? We cannot simply figure out the kinetic energy of the bullet, since the wall did more than just stop it, it flattened it. $\endgroup$– 228Commented Jul 8, 2021 at 2:54
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2$\begingroup$ @228 the wall did exert a force on the bullet, and the bullet did accelerate, since it went from whatever speed it originally had to $v = 0$. $\endgroup$– AllureCommented Jul 8, 2021 at 2:55
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$\begingroup$ @Allure Wouldn't that be decelerating? $\endgroup$– 228Commented Jul 8, 2021 at 3:03
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2$\begingroup$ @228 in physics, acceleration is the same as deceleration, it just has a negative sign: e.g. if you decelerate at $5 m/s^2$ that's the same as accelerating at $-5 m/s^2$. $\endgroup$– AllureCommented Jul 8, 2021 at 3:13
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3$\begingroup$ @228 Allure is right. In physics, acceleration is any change in velocity. We just use the term acceleration, whether the speed is increasing, decreasing, or even remaining constant (direction changes also count as accelerations, since the velocity vector is changing). $\endgroup$ Commented Jul 8, 2021 at 3:38
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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 net 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.
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$\begingroup$ What happens if we say that a force does work even if the material at the point of application is stationary? Do we all float off into space and die? Seriously, I mean, what happens if we view it from that perspective? What can we no longer make sense of? I understand you would have to change some definitions, but it seems that definitions already have multiple meanings depending on different variables. I mean, surely a force that does no work and a force that does work are distinctly different, but they are still both called forces. $\endgroup$– 228Commented Jul 10, 2021 at 5:37
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$\begingroup$ @228 said “What happens if we say that a force does work even if the material at the point of application is stationary?” Then work would not be equal to the transfer of mechanical energy. Since the transfer of mechanical energy is an important concept we would use some other word and some other variable to refer to $\vec F \cdot \vec d$. $\endgroup$– DaleCommented Jul 11, 2021 at 0:57
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$\begingroup$ @228 also said “a force that does no work and a force that does work are distinctly different, but they are still both called forces”. The thing that makes a force a force is that it transfers momentum. Some forces also transfer energy, but that is not what defines a force. A force is a force because it transfers momentum. Work is a transfer of energy. Both momentum and energy are important concepts, so we are interested in how each of them transfer. $\endgroup$– DaleCommented Jul 11, 2021 at 1:02
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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.
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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.
