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Figure 1: Work done on a block by a constant force of \(-10\) newtons over a displacement of \(5\) meters.
The work \(W\) (measured in joules, the unit of energy) done on an object by any force \(\vec{F}\) essentially measures how much energy \(E\) that force either transferred into or out of that object. If \(W\) is positive, then the force \(\vec{F}\) put energy into the object; if \(W\) is negative, then the force \(\vec{F}\) took energy out of the object.
Let's look at a few examples to see what I mean by this. Suppose that a block of mass \(m=5kg\) is moving to the right at a constant velocity of \(v=+30m/s\). The total energy \(E\) of the block is just its kinetic energy; thus, \(E=KE=\frac{1}{2}(5kg)(30m/s)^2=2,250J\). Suppose that a constant force \(F_A=-10N\) (acting in the left direction as depicted in Figure 1) is applied to the object as it is displaced by an amount \(Δx=10m-5m=5m\) as depicted in Figure 1. (We'll assume that the force \(\vec{F}_A\) does not act on the object for \(x\) positions of \(x<5m\) and \(x>10m\); in other words, \(F_A\) acts on the object only during the time interval when it moves from \(x=5m\) to \(x=10m\).) The work done on the object by \(F_A\) is given by
$$W=(-10N)(5m)=-50J.\tag{1}$$
As we discussed earlier, if the work done by a force on an object is negative, then that force took energy out of the system. Equation (1) tells us the following: the object initially had a lot of energy (\(E_i=KE_i= 2,250J\)), but the force \(F_A\) did a negative amount of work on the object (\(W=-50J\)) and transferred \(100J\) of energy out of the object; consequently, the object has a lower amount of energy given by \(E_f=E_i+W= 2,250J-50j=2,200J\). What is the physical interpretation of this object's loss of energy? We are assuming that this object is a block of constant mass whose height above the ground doesn't change; thus, neither \(m\) nor \(PE\) change. The only thing that could have changed about the object is its velocity. Using algebra and the conservation of energy, we can find out how much the object's velocity changed by:
$$ΔE=W$$
$$E_f=E_i+W⇒\frac{1}{2}mv_f^2=\frac{1}{2}mv_i^2+W$$
$$⇒v_f=\sqrt{v_i^2+\frac{2W}{m}}= \sqrt{(30m/s)^2+\frac{2(-50J)}{5kg}}=\sqrt{(400m/s)^2}=20m/s$$
The goal of physics is to say something "concrete" about the world; but saying that an object's energy changed from \(2,250J\) to \(2,200J\), alone, without saying anything else, isn't really saying anything "concrete" about what happened (at least on a superficial level) since energy is a very abstract notion. But that's only on the surface. The nice thing about physics is that the abstract quantities which we deal with (i.e. energy) are, despite on a superficial level being very abstract, actually predicting something or "telling us" something about the nature of the world in a way that is concrete. To answer the question I posed earlier, the physical interpretration of the object's loss of energy is that the force \(F_A\) must have had slowed the object down from an initial velocity of \(+30m/s\) to a final velocity of \(+20m/s\).
Why is it that if a force does negative work \(W\) on an object, that object loses an amount of energy given by \(W\)? The short answer is, conservation of energy, which I have written below
$$ΔKE+ΔPE=W+Q+T_{other}.$$
The left-hand side of the equation is the object's change in total energy. The right-hand side is essentially a list of all the possible different ways an object can gain or lose energy: doing work \(W\) on the object, the object getting heated or cooled by an amount \(Q\), or \(T_{other}\) which refers to other ways such as, for example, transferring matter into the object or striking the object with X-rays. In our simplified example, all of the terms on the left- and right-hand sides cancel except for \(ΔKE\) and \(W\) leaving us with just
$$ΔKE=W.$$
What if a force \(F_B=+10N\) acted on the object to the right? What would be the work done by \(F_B\)? What would the physical interpretation be of positive work transferring energy into the object? Essentially, to answer these questions, we would go through all the same steps as in the previous example except replace \(F_A=-10N\) with \(F_B=+10N\). I leave it as an exercise for the reader to substitute \(F_B\) for \(F_A\) to answer these questions.
Sources: Khan Academy