LURCH said:
So then, you're saying the acceleration takes place without work being done, right?
No.
Please go back and read what I actually wrote. (You should also learn the difference between a logically necessary condition and a logically sufficient condition.)
"Force," "Work," "Velocity," and "Acceleration" are all
physically distinct concepts, as should be obvious from the mere fact that
they each have different physical units.
It is
possible to accelerate an object and do work, if the object's speed changes. It is also
possible to accelerate an object
without doing work, by changing the direction of its motion
without changing its speed. And it is also possible to do work
without changing the speed of any macroscopic body, e.g., by exerting force on a dashpot piston. However,
none of these conditions are
necessary conditions; they are merely
possible conditions --- because
Force, Work, Velocity, and Acceleration are all physically distinct concepts.
Consider the following explicit example of the second case. The rate of work on an object is given by dW/dt =
F.v, where work W is a scalar, the force
F and the velocity
v are vectors, and `.' denotes the "dot product" between vectors. From Newton's 2nd Law,
F = m
a; therefore, dW/dt = m
a.v. Expanding the dot product in a cartesian coordinate system, dW/dt = m(a_x * v_x + a_y * v_y + a_z * v_z). Consider the case where a unit force parallel to the Z-axis is applied to a unit mass. The velocity vector and the acceleration vector are physically distinct quantities. Consider the case where the velocity is instantaneously along the X-axis. Then v_x = 1 m/s, v_y = 0 m/s, v_z = 0 m/s, while a_x = 0 m/s^2, a_y = 0 m/s^2, a_z = 1 m/s^2. The velocity is not zero. The acceleration is not zero. The acceleration is perpendicular to the velocity. Thus, the dot product of velocity and acceleration is zero:
F.v = m*[(1 m/s)*(0 m/s^2) + (0 m/s)*(0 m/s^2) + (0 m/s)*(1 m/s^2)] == 0 kg*m^2/sec^3.
Therefore, the rate of work is zero in this specific case, even though neither the force, nor the velocity, nor the acceleration vanish.
Contrawise, consider a force exerted on a linear dashpot piston oriented along the X-axis. In this case, F_x = k * v_x, where `k' is the viscous friction coefficient of the dashpot. If the piston is moving at a constant speed equal to the terminal velocity for the applied force, the acceleration is zero. The constant rate of work, however is dW/dt = k * (v_x)^2, which is
not in general zero.
Thus, one has seen an explicit case where acceleration occurs without work, and an explicit case where work occurs without acceleration.
Again,
Work, Force, Velocity, and Acceleration are all physically distinct concepts, as evidenced by the fact that they each have different physical units. It is
possible (but not necessary!) to have acceleration without doing work, and it is
possible (but not necessary!) to do work without having an acceleration, because
Work and Acceleration are physically distinct concepts; neither one necessitates the other, as has been explicitly shown through two physical examples.