Does greater acceleration result in more work done?

[Imperial Sky]

Work W done by moving the object with force F for distance s is W = Fs.
When I move the same object the same distance but with twice the acceleration, does
the work done gets also doubled?
By F=ma, doubling the acceleration yields m*2*a = 2F -> 2Fs = 2W.

I've mostly read, that if I want to increase work, I must either increase the mass of the object or
the distance, which gets me confused, because acceleration is also in the equation.

 

[Merlin3189]

I think your problem is saying that the same force can cause different acceleration.
As you say, F=ma, so to get double the acceleration you need twice the force.

I've mostly read, that if I want to increase work, I must either increase the mass of the object or
the distance, which gets me confused, because acceleration is also in the equation.

Well I wouldn't agree with that. Your original definition is the right one W=Fs so to increase work you must increase the force or the distance through which it moves.

Say you push a 5 kg mass with a force of 10 N over a distance of 3 m, then your work is 30 J.
If you treble the mass to 15 kg, but keep the force to 10 N, then over 3 m you still do 30 J of work.
The difference is that now the acceleration will be less and it will take you longer (time) to move through those 3 m.

You have to be clear about what changes you are making and ensure the changes you make are possible.
For example, you can't change acceleration but keep both force and mass the same, because F=ma. You can't change work if you keep both force and distance the same, because W=Fs.

 

[Imperial Sky]

If I move a 1kg body with an acceleration of 1 m/s^2 (F = ma = 1N) for 1 meter, I do 1J work.
If I move the same 1kg body the same 1 meter distance, but with acceleration 2 m/s^2 (F=2N), does that mean that I did 2 times more work (2J) than before?

 

[Merlin3189]

Yes.
W = Fs = 2N x 1m = 2 J

 

[A.T.]

If I move a 1kg body with an acceleration of 1 m/s^2 (F = ma = 1N) for 1 meter, I do 1J work.
If I move the same 1kg body the same 1 meter distance, but with acceleration 2 m/s^2 (F=2N), does that mean that I did 2 times more work (2J) than before?

How much kinetic energy will the object gain in either scenario, if it starts from rest?

 

[Merlin3189]

In the first case, the work done is 1 J so it gains 1 J of KE if there is no friction etc.
In the second case, the work done is 2 J so it gains 2 J of KE if there is no friction etc.

You can check with your equations of motion:
1)$a= 1 m/sec^2 \ \ \ ⇒ v^2=2as = 2 \times 1 \times 1 = 2 m^2/sec^2 \ \ so \ \ KE = \frac {mv^2}{2} = \frac {1\times 2}{2} =1 J$
2)$a= 2 m/sec^2 \ \ \ ⇒ v^2=2as = 2 \times 2 \times 1 = 4 m^2/sec^2 \ \ so \ \ KE = \frac {mv^2}{2} = \frac {1\times 4}{2} =2 J$

 

[Nugatory]

I've mostly read, that if I want to increase work, I must either increase the mass of the object or
the distance

That's assuming that the object starts at rest and ends at rest, as if for example we're lifting it off the floor and setting it on a table. The work done is $mgh$, and $g$ is constant so to increase the work done we need either a heavier object or a higher table.

If the object is moving at a different speed when we're done, we also have to allow for the work done to change its kinetic energy. In your problem, the greater force over the same distance means a greater final speed and hence greater kinetic energy.