Since $W=Fs$, $F=\frac{W}{s}$. When you substitute this in the formula for acceleration, $a=\frac{F}{m}$, you will get that $a=\frac{W}{ms}$. Then, when work equals zero, acceleration will be zero.
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14$\begingroup$ $W ≠ FS$ but $W = \int F.dS$ $\endgroup$– An_ElephantCommented Jan 26, 2023 at 17:25
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1$\begingroup$ How could there be? Don't you see acceleration as some kind of change? How does the First Law of Motion allow change without work? $\endgroup$– Robbie GoodwinCommented Jan 27, 2023 at 19:20
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1$\begingroup$ @RobbieGoodwin: Take a look at the top rated answer for how you can have acceleration without work. $\endgroup$– Michael SeifertCommented Jan 27, 2023 at 22:01
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1$\begingroup$ @MichaelSeifert Why not try to explain that? $\endgroup$– Robbie GoodwinCommented Jan 27, 2023 at 22:14
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2$\begingroup$ It is the "dot" of the "dot product" that is important. Even if we consider constant forces and constant displacements, the work is not $W=FS$ but $W=\vec F\cdot\vec S$. If the angle between $\vec F$ and $\vec S$ is $\theta$, then we have $W = FS\cos\theta$. If the force and the displacement are perpendicular ($\theta=90^\circ$), the work is zero. $\endgroup$– printfCommented Jan 28, 2023 at 3:17
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7 Answers
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Can there be acceleration without work?
Yes. An object going round in a horizontal circle at constant speed is accelerating because the direction of its velocity is changing. However, the magnitude of its velocity (its speed) is constant, and so its potential and kinetic energy are constant. Therefore it neither does work nor has work done on it. Although there is a force acting on the object (the centripetal force which keeps it moving in a circle) this force is always at right angles to the velocity of the object and so it does no work on the object.
answered Jan 26, 2023 at 8:59
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1$\begingroup$ Kinetic energy is transfer of energy, so work is doing. What is potential energy here. $\endgroup$ Commented Feb 4, 2023 at 14:14
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$\begingroup$ By definition, no. What can 'acceleration' be but a change…? Who doubts a circling object is accelerating, because its direction is changing? If in velocity, only speed mattered, potential and kinetic energy could only be constant with continuous input to at least one… if it's moving, neither kind of energy is constant. Can you explain how a force at right angles to the velocity of an object 'does no work…'? Does gravity 'do no work' on objects in orbit? $\endgroup$ Commented Feb 6, 2023 at 22:03
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1$\begingroup$ @RobbieGoodwin "if it's moving, neither kind of energy is constant" - incorrect. An object moving along a horizontal path (e.g. a horizontal circle) at constant speed has both constant kinetic energy (because its speed is constant) and constant potential energy (because the path is horizontal). However, it may still be accelerating if the direction of its path changes. "Does gravity 'do no work' on objects in orbit?" - yes, if the orbit is circular so KE and PE are both constant. $\endgroup$ Commented Feb 7, 2023 at 9:40
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$\begingroup$ Of course the object moving in a circle at constant speed is accelerating because the direction of its velocity is changing. That doesn't alter the fact that the velocity is being changed by a force. Whether that's 'pure' gravity, or some centrifugal or centriputal of G, I don't mind. Either that, or Newton was wrong. Which would you prefer? $\endgroup$ Commented Feb 13, 2023 at 19:05
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$\begingroup$ @Gandalf How would you you define that 'horizontal circle' please? $\endgroup$ Commented Feb 13, 2023 at 20:30
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Yes, there can be acceleration without work.
Consider a particle on a string in uniform circular motion with constant speed. Since the speed is constant the kinetic energy (KE) is constant. But due to the centripetal force on the particle from the tension in the string, the particle has acceleration toward the center of the circle. The work from the tension force is determined by $\vec T \cdot \vec v$ and since $\vec T$ and $\vec v$ are at 90 degrees there is no work. The velocity vector changes, hence the particle is accelerated; the speed (magnitude of the velocity) is constant, hence the KE is constant and there is not work done on the particle.
answered Jan 26, 2023 at 9:09
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2$\begingroup$ Centripetal force is for frame not particle. $\endgroup$ Commented Jan 26, 2023 at 9:41
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6$\begingroup$ @NeilLibertine I'm not sure what you mean. Centripetal force is a force that acts on the particle. It's a real physical force - if it's too strong it will break the string. In a rotating reference frame there is an additional fictional force, centrifugal, that acts in the opposite direction and stops the particle accelerating. $\endgroup$– bdslCommented Jan 27, 2023 at 17:02
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3$\begingroup$ Centripetal force is a true force – it is the actual physical force that makes an object move in a circle. For a mass on a string it is the tension force in the string, pulling the mass inwards. For a gravitational motion of a small body (e.g. Earth) around a much larger body (e.g. Sun) it is the gravitational force, again pulling the body inwards. $\endgroup$– printfCommented Jan 28, 2023 at 2:54
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Magnetic forces don't do work on an otherwise free particle because the force is always perpendicular to the velocity. Without using calculus, you have to remember that the work $w=F S$ only when $F$ and $S$ are parallel. More completely, $w=F S \cos(\theta)$ where theta is the angle between $F$ and $S$. In the magnetic case, $\theta=90$ degrees so $\cos\theta=0$ and $w=0$
With calculus, the force is defined as $\vec{F}=q\,\vec{v}\times\vec{B}$ where $\vec{v}$ and $\vec{B}$ are the velocity and magnetic field. A useful fact to know is that the cross product of two vectors will always be perpendicular to the original two vectors. Now calculate the work, $w=\int \vec{F}\cdot d\vec{s} = \int (q\,\vec{v}\times\vec{B})\cdot d\vec{s}$. Recall that $d\vec{s} = \vec{v}\,dt$ so $w=\int (q\,\vec{v}\times\vec{B})\cdot \vec{v}\,dt$. The useful fact above says that the dot product in the integrand is 0 so the work is 0.
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The Coriolis force is always orthogonal to the velocity vector so it does not do any work on a fluid parcel it acts on, but it causes acceleration that changes the direction of the velocity vector.
When deriving an equation for the kinetic energy balance (or turbulent kinetic energy balance), the Coriolis term drops out.
answered Jan 27, 2023 at 9:48
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The fundamental physics principles state that energy does not vanish, it just changes its form. Consider E=mc^2, electromagnetic waves interacting on particles or circular motion.
Work is the change in energy (e.g. difference of potentials energies).
Acceleration is due to the change in velocity and thus change in kinetic energy.
When you have acceleration you increase kinetic energy and this reduces other types of energy (or similar resource). Therefore you cannot have acceleration without work.
Most of the answers claiming that circular motion has acceleration without work are neglecting the fact that the object creating the force at the center is also affected by the object circulating the center (e.g. Moon creates tides on Earth).
The simplest example is a Olympic sport of hammer throwing. The thrower stands straight initially straight. Once there is motion in the hammer, the thrower leans in the opposite direction to gain balance and does work into the system.
For example, a 100kg person is moving his mass by 10 cm to balance the centripetal force of the hammer. Thus, in radial direction only, the hammer thrower has done work W = F.s where F is the horizontal Force component caused by the gravity and the tilt-angle of the thrower and s is the horizontal distance of the tilt. The F equals the centripetal force of the hammer (and the corresponding centripetal acceleration). The work has to come from somewhere. The same applies to all fields (magnetic, gravity, etc.).
No Work, No Acceleration.
Don't believe everything you are told to.
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1$\begingroup$ I guess I am just repeating myself at this point, but this is fundamentally wrong: "Acceleration is due to the change in velocity and thus change in kinetic energy." Velocity is a vector. Acceleration is a vector. Change in velocity does not imply change in kinetic energy. The circular motion due to a centripetal force (e.g. gravitational force) towards a fixed centre is an example. The velocity (a vector) is constantly changing, thus there is acceleration. The speed (absolute value of velocity) does not change, hence the kinetic energy ($\frac12mv^2$) does not change. $\endgroup$– printfCommented Feb 7, 2023 at 2:57
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$\begingroup$ @printf There does not exist "a fixed centre" anywhere in the known universe. This is an imaginary object. The rotating object always affects the central object and there is motion of these both. This creates work. The "centripetal story" assumes that there are fixed objects which is not true in the real world. $\endgroup$– JuhaCommented Mar 8, 2023 at 22:12
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$\begingroup$ Nobody is talking about a "fixed centre". The rotation of the Earth around the Sun – or the rotation of both around the common centre of mass – will continue forever (at least in classical physics, ignoring gravitational radiation). No work is done. No energy is expended. $\endgroup$– printfCommented Mar 10, 2023 at 5:46
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Simple answer is Gravity.
When a ball dropped from a height, it experienced acceleration without any external work or force acting on it.
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2$\begingroup$ This is deeply incorrect. In newtonian mechanics (the context of the question), gravity leads to an external force… It's called weight, or gravitational force, and it does provide the work that makes the object accelerate during its fall. $\endgroup$– MiyaseCommented Jan 29, 2023 at 11:43
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$\begingroup$ Obviously this is very incorrect. It is the gravitational force that causes the ball to accelerate towards the Earth, and it certainly does work. $\endgroup$– printfCommented Jan 29, 2023 at 23:15
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This depends on the definition of work.
The answers including circular trajectory should also include the mention that acceleration towards center has a change of distance towards the center. Thus, if you define work as force times travelled distance, acceleration towards center also includes work.
However this work is immediately released to the change of motion from linear path.
The discrepancy here is that usually definition of work is taken from linear systems that is not compatible with particles in nonlinear systems. Definition of acceleration and work have to be compatible.
In other words:
No, there cannot be acceleration without work. Otherwise you violate Newton's laws and you get a Nobel price.
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1$\begingroup$ "If you define work as force times travelled distance" – work is defined as (an integral of) the dot product of force (a vector) and displacement (a vector). If these vectors are orthogonal (perpendicular), the work is zero. (And hence the change in kinetic energy is zero.) If a body moves in circular trajectory (with fixed radius) around another body, then the work performed is zero, and the change of kinetic energy is zero, and this circular motion can continue forever. $\endgroup$– printfCommented Jan 28, 2023 at 3:04
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$\begingroup$ @printf yes, in a circular track there is displacement towards center (displacement =>acceleration) and a force towards center (keep particle in track). Hence, there is work towards center. There is a problem with the definitions. The definition of work is different in circular track or definition of acceleration is different (not towards center or not at all). You cannot take one from linear coordinate system and the other from circular. In other words, the has to be both acceleration and force (F=ma) => work. $\endgroup$– JuhaCommented Feb 2, 2023 at 7:40
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$\begingroup$ @printf They are NOT orthogonal. Centripetal force and centripetal acceleration are parallel. That is my whole point. If you accept that centripetal acceleration is real acceleration you have to accept that there is displacement radially as acceleration is the second derivative of displacement. The particle is kept in track by a centripetal force. Hence, there is force and displacement radially. The problem here is that centripetal force and acceleration are not real but mathematical constructions. Thus, No work, no acceleration. $\endgroup$– JuhaCommented Feb 4, 2023 at 11:10
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$\begingroup$ The point is that if a body travels in a perfect circle (not a spiral), then the displacement is always orthogonal to the force applied (and to the corresponding centripetal acceleration), and hence there is no work. The sun does not do any work on earth (in the model where the earth rotates around the centre of the sun). The earth does not come any closer to the sun; it stays at the same distance. There is force and acceleration; there is no work. The sun does not spend any energy. $\endgroup$– printfCommented Feb 5, 2023 at 1:09