I was wondering if I could get some help closing some fundamental gaps in my intuition of work, as it relates to force and distance travelled.


Say we pull a 1kg box along the ground. We pull with a force of 1N parallel to the ground, over a distance of 1 meter. We have a formula which gives us work done as $$W = F\cdot d \cdot \cos\alpha = 1N \cdot 1m \cdot 1 = 1J$$

Now, say we pull the box along the ground with the same force, over the same distance, but we pull at an angle of $\alpha = 60^\circ$. Now, the work we've done is $$W = 1N \cdot 1m \cdot \cos60^\circ = 0.5J$$

So, this is where my intuition breaks down a bit, and I think it is because of a warped understanding of work. I would have expected that while only 0.5J was spent moving the box horizontally, the remaining 0.5J dissipated somewhere vertically. Was it spent acting against gravity, and gravity won?

My thoughts

As a surface level understanding, I would say that the work done can either be a product of force and distance, but it can also be seen as the change in energy states of the box.

For instance, if I lifted the box 1m straight up by applying an arbitrary force straight up, I would have caused its potential energy to rise by 9.81J. But I have also moved it 1m upwards, so I have also done some more (?) work that would be given by the product of force and distance again. It is my understanding that we don't "add" these two calculations up to find total work done, but rather they are two different calculations that yield the same result?

But the work I do using the $W = Fd$ formula (assuming the force is parallel to the travel vector) can vary wildly depending on how much force I apply to the box, right? And if so, how can it always coincide with the change in the box' potential energy?

My questions

  1. When we move the box horizontally, we don't change its potential energy. I'm presuming (and this is what I hope to clarify) that if we measured the kinetic energy of the box over time, it would somehow yield the 1J again, even though it starts and ends at zero. Do we just consider the highest kinetic energy change, and that yields 1J?

  2. In the case of vertical movement, we change the box' potential energy which is a measure of how much work we applied to it. But since we also moved the box over a distance, we have two different formulae, and I'm having a hard time consolidating them. If we consider $\Delta E_p = mgh_2 - mgh_1 = 9.81J$ always, and $W = Fd$ which varies depending on the magnitude of the force, how can they always be the same?

I see how since $F=mg$, the latter equation "becomes" the former, so I'm not blind to the mathematical obviousness. As I mentioned early on, my question is on the intuition.

  • $\begingroup$ Draw a free body diagram for each case. Also, does the box weigh more or less than 0.866 N? $\endgroup$ Mar 13, 2023 at 23:55
  • $\begingroup$ @DavidWhite - Say the box weighs 1kg, it would weigh more than 0.866N. I assume you're considering $\sin60^\circ$ here? $\endgroup$
    – Alec
    Mar 13, 2023 at 23:58
  • $\begingroup$ Alec, indeed I am. A 1 kg box weighs 9.8 N at sea level. I'll attempt to answer your question based on a 1 kg box. $\endgroup$ Mar 14, 2023 at 0:03
  • $\begingroup$ It is helpful to remember that the work of the NET force equals the change of the kinetic energy. In the first case the net force is the difference between applied and friction horizontal forces. In the second case, the difference between applied upward force and the weight $mg$. $\endgroup$ Mar 14, 2023 at 0:55
  • $\begingroup$ what do you mean it [box's kinetic energy] starts and ends at zero? If you do 1J of work pulling it, it will have 1J of KE. $\endgroup$
    – JEB
    Mar 14, 2023 at 3:18

3 Answers 3


The most important things correlated with work is which force does work. The net work is the algebraic sum of all the work, either positive work or negative work. When you are analyzing work, you only need to first analyze the force exerted on the object and the direction of movement and don't care about others.

Your scenario is two different situations, so you will have two different energy levels.

The change of kinetic energy equals to the net work done but not the work done by net force.


When an object moves, the component of a force in the direction (or opposite the direction) of the object can increase or decrease the velocity and thus the kinetic energy of the object. Force perpendicular to the motion can only cause the velocity to change direction of the velocity. Although this affects the velocity, it does not increase or decrease it. The perpendicular component cannot change the kinetic energy. The perpendicular force is not lost, but it does not do work. Energy is not a vector. It does not have direction.

In your example, gravity pulls down and a normal force pushes up to prevent the object from moving through the floor. When you pull at an angle, the vertical component pulls against gravitational force. This reduces the normal force needed, so the perpendicular component does have an effect. If the block feels friction, then this reduced normal force results in reduced friction.

As for potential energy, lifting a block does positive work on the block because it pulls in the forward direction. Gravitational force pulls against the motion and does negative work. Because gravity is a conservative force, the gravity can later do positive work on the block as it falls back down down. This works as if energy is taken from the block as it rises, stored as potential energy in the force, and then given back as kinetic energy at a later time. This model allows for the method of energy conservation without having to calculate work done by conservative forces. Some view potential energy as a real form of mechanical energy. Some view it as a way to bypass calculating work for forces that are not constant.


Assuming that the box has a mass of 1 kg, weighs 9.8 N (i.e., on the surface of the earth), and is on a frictionless horizontal surface, a 1 N force can't lift the box off of that surface. For the case where the 1 N force is pulling the box horizontally, the x-component of that force is 1 N and the y-component of that force is 0 N, so all of the force is involved in accelerating the box and increasing its kinetic energy, per the work-kinetic-energy theorem. For the 60 degree angle relative to horizontal, the x-component of the force is 0.5 N and the y-component of that force is 0.866 N. The box is too heavy to accelerate vertically, and indeed, only the component of the force that is parallel to the displacement is involved in work. Thus, 0.5 N of force accelerates the box in a horizontal direction, resulting in less work than in the first case.

  • $\begingroup$ Ok, this seems to look at the case of horizontally moving the box. But I don't see how this answers my overarching question of how to consolidate understanding work as a force*distance result, and as a "change in energy level" result. $\endgroup$
    – Alec
    Mar 14, 2023 at 0:33
  • $\begingroup$ @Alec, you may be missing a key concept. In physics, work is defined such that ONLY the component of a force that is parallel to the displacement adds or subtracts to or from the work done. If the box is lifted straight up, it will gain gravitational potential energy and kinetic energy. If the box slides sideways, it will gain only kinetic energy. $\endgroup$ Mar 14, 2023 at 16:37

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