Say I push a box with a constant force $F$ causing a displacement $d$. So magnitude of work done by me on the box is $W=Fd$, and the kinetic energy gained by the box is $W$. But according to N3L, the box exerts an equal and opposite force $F$ on me, but since I am moving opposite to direction of force exerted by box, the box does $-W$ work on me, but does this mean that the box gains even more work $W$, and I am losing more energy equal to $W$? But that is paradoxical as I only applied force $F$ along displacement $d, and my interpretation leads to me constantly losing energy and unintentionally doing more work because of N3L. How does this make sense? How should I treat Newton's third law in this context?
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asked Apr 11, 2022 at 18:49
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$\begingroup$ It is not paradoxical. There are no paradoxes, only apparent paradoxes. In this case, you have to consider that there are other forces on you than just the force of the box on you. The force of the box on you is equal in magnitude and opposite in direction to the force of you on the box, so if that was the only force on you, you would start accelerating away from the box. But you seem to imply that you are moving in the same direction as the box, so either you started off with some non-zero velocity in the direction of the box motion, or there are other forces on you. $\endgroup$– hftCommented Apr 11, 2022 at 18:59
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$\begingroup$ Thanks @hft, got it $\endgroup$– AltercatingCurrentCommented Apr 11, 2022 at 19:19
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2 Answers
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Only forces on a body determine the work done on the body. In your case, work done on the box depends on the force applied to the box by you. For your case, the reaction force is not on the box, it is on you and has nothing to do with the work done on the box. The work done on you depends on all the forces acting on you, including the force from the ground on you when you push the box. Work is not stored in the system, it is energy applied that changes the kinetic energy of the system.
answered Apr 11, 2022 at 19:06
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$\begingroup$ "Only forces on a body determine the work done on the body." I understand that, but my confusion lies in the fact that does it mean that work done on a different body doesn't mean energy loss? Like if the box applies negative work on me, it means that I'm losing energy, but does it also not mean that I'm losing energy when I'm doing positive work on the box? $\endgroup$ Commented Apr 12, 2022 at 3:26
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$\begingroup$ My confusion is arising believing that work done by a body = energy lost by the body and treating action and reaction as two different forces, both of which I assume are true? $\endgroup$ Commented Apr 12, 2022 at 3:32
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1$\begingroup$ Here we deal with point body or rigid body for which internal energy of the body cannot change, work from a force acting through a distance can only change the kinetic energy (KE). Negative work is decrease in KE. (Potential energy just a convenient way to evaluate work from conservative force. In general, bodies have internal energy and work can change internal energy; e.g., gas expanding against a piston. This is subject of thermodynamics which broadens simple mechanics definition of work. See my answer physics.stackexchange.com/questions/703246/… $\endgroup$ Commented Apr 12, 2022 at 12:11
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$\begingroup$ Remember change in KE is from all forces. When you push a body the reaction force on you does - work but the force of ground on you does + work and unless you gain KE net work done on you is zero. $\endgroup$ Commented Apr 12, 2022 at 12:31
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You do work W on the box and the box does work -W on you. That much is true. It is also true that the total work done on the system consisting of you plus box is $W_{total}=W+(-W)=0$.
In some respects work and energy resemble money. Say you borrow \$10 from me. If your reasoning holds, then I should expect you to pay me back \$20.
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$\begingroup$ I don't think the money analogy is helping in my case, I've asked some more questions in the comments section under John Darby's answer, you might want to check that $\endgroup$ Commented Apr 12, 2022 at 3:35
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$\begingroup$ Forces mediate the transfer of energy by doing work. Consider a collision between two balls moving in the same direction, the one in the back moving faster than the one in the front. While the balls are in contact, they exert action-reaction forces on each other. The ball in the back does positive work on the ball in the front and at the same time the ball in the front does negative work on the ball in the back. The ball in the back loses an amount of kinetic energy equal to W while the ball in the front gains that same amount of kinetic energy. $\endgroup$– AdHocCommented Apr 12, 2022 at 4:17