Why isn't work relative? Acceleration and displacement can be defined relative inertial frames. For example, a body moving with constant velocity on top of another moving with the same velocity will have zero displacement with respect to latter body. Similarly, two bodies in free fall will have zero acceleration with respect to other.
o work is defined as $$W=\int_{x_1}^{x_2}\vec{F}.\vec{ds}$$ or
$$W=\int_{t_1}^{t_2}\vec{F}.\vec{v}\,dt$$
In both of these forces, velocities and displacements can be calculated using different inertial frames as their references. So why isn't work relative?
 A: Work is a definite rather than an indefinite integral: $W = \int_{\,\vec x_0}^{\,\vec x_1} \vec F \cdot d\vec s = \int_{t_0}^{t_1} \vec F \cdot \vec v\,dt$. Both forms strongly suggest that work is a frame-dependent quantity, and it is. You have to be very careful to stick with one frame of reference when using frame-dependent quantities such as velocity, energy, or work. You're likely to get sheer nonsense if you mix and match frames of reference.
Here's a very simple example. Suppose a constant force of one newton is applied for two seconds to an object with a mass of one kilogram. This force changes the object's velocity by two meters per second. The work done on the object is


*

*2 joules from the perspective of a frame in which the object in question starts at rest,

*-2 joules from the perspective of a frame in which the object comes to rest at the end of the interval, and

*Zero from the perspective of a frame in which the object instantaneously comes to rest in the middle of that interval.
A: The work done is relative to your inertial reference frame, as are the kinetic energies. But of course everything is consistent in the end - let's verify that in an example:
Take a mass accelerated from $v_1$ to $v_2$ by a force $\vec F$ as measured in an inertial reference frame $R$. The change in kinetic energy and work are related by:
(1)  $\triangle KE = \frac{1}{2}m(v_2^2 - v_1^2) = \int_{t_1}^{t_2} \vec F \cdot \vec v \space dt = W$
Now look at the same situation from a reference frame $R'$ moving at velocity $-\vec v_r$ relative to $R$:
(2)  $\triangle KE' = \frac{1}{2}m(v_2'^2 - v_1'^2) = \int_{t_1}^{t_2} \vec F \cdot \vec v' \space dt = W'$
We want to show that this equation is still true in frame $R'$. The velocities in the two frames are related by $v_i' = v_i + v_r$:
(3)  $\frac{1}{2}m((\vec v_2+\vec v_r)^2 - (\vec v_1+\vec v_r)^2) = \int_{t_1}^{t_2} \vec F \cdot (\vec v + \vec v_r) \space dt $
(4)  $\frac{1}{2}m(v_2^2 - v_1^2 + 2\vec v_r \cdot (\vec v_2-\vec v_1)) = \vec v_r \cdot \int_{t_1}^{t_2} \vec F \space dt + \int_{t_1}^{t_2} \vec F \cdot \vec v \space dt $
Subtract equation (1) from equation (4) to get
(5) $\vec v_r \cdot m(\vec v_2 - \vec v_1) = \vec v_r \cdot \int_{t_1}^{t_2} \vec F dt$
Replace $\vec F=m\vec a$ and divide by $m$, :
(6) $\vec v_r \cdot (\vec v_2 - \vec v_1) = \vec v_r \cdot \int_{t_1}^{t_2} \vec a \space dt$
Which is obviously true since $(\vec v_2 - \vec v_1) = \int_{t_1}^{t_2} \vec a \space dt$.
Edit: I'm still a bit short on reputation to participate in the commentary discussion, so I'll note here that the line elements $ds$ and $ds'$ are not equal. If you identify the integral paths $s(t)$ and $s'(t)$ as being the same physical path in two different frames, then in one dimension we have:
$\frac{ds'}{ds} = \frac{ds'}{dt} / \frac{ds}{dt} = \frac{v'}{v} = \frac{v + v_r}{v}$
therefore $ds' = (1 + v_r/v) ds$
