Is Infinite Work Possible in a Frictionless Vacuum?

[suchal]

Work done= force.displacement
In space, with no external forces, air drag, gravity etc if you apply a force to object if will move forever in the direction of force unless any resultant force act on it to change its momentum.
In this case let's take force as 2N so we get
w.d=2N.s
s will increase forever so w.d is infinite?
I know i am wrong so please help me get the concept right.

 

[CompuChip]

Hi,

The "s" in the formula is the distance ovet which the force acts, not the distance that the object will travel /after/ the force has been applied.

 

[suchal]

Hi,

The "s" in the formula is the distance ovet which the force acts, not the distance that the object will travel /after/ the force has been applied.

Sorry but i still don't get it. You mean as distance which the object travels while being accelerated by force?

 

[Dale]

The mathematical definition is W=\int \mathbf{f}\cdot \mathbf{v}\;dt So for all of the time that f=0 you have W=0 even though v≠0.

 

[CompuChip]

What Dale said :-)

To give a numerical example, suppose that I push a block at a constant force of F = 10 N. Initially the block is at rest, and after 2 seconds I stop pushing. The total work on the block is $W = (10~\mathrm N) \cdot (2~\mathrm s) = 20~\mathrm J$. After this there will be no force, so the additional amount of work on the block over any distance s is $W = 0 \cdot s = 0$. The block will continue moving at its final speed (which follows from $20~\mathrm{J} = \tfrac12 m v_\mathrm{final}^2$ where m is the mass of the block).

 

[sophiecentaur]

Work done= force.displacement
In space, with no external forces, air drag, gravity etc if you apply a force to object if will move forever in the direction of force unless any resultant force act on it to change its momentum.
In this case let's take force as 2N so we get
w.d=2N.s
s will increase forever so w.d is infinite?
I know i am wrong so please help me get the concept right.

The answer to any question about a real situation is never "infinity". If you are applying the Force all the time then the work done will 'tend to' infinity. But you don't have long enough (or enough energy) to do the experiment for "for ever".
Along with Voltage Sources and Short circuits, the original model needs qualification if you want a proper answer from any question about it.

 

[suchal]

What Dale said :-)

To give a numerical example, suppose that I push a block at a constant force of F = 10 N. Initially the block is at rest, and after 2 seconds I stop pushing. The total work on the block is $W = (10~\mathrm N) \cdot (2~\mathrm s) = 20~\mathrm J$. After this there will be no force, so the additional amount of work on the block over any distance s is $W = 0 \cdot s = 0$. The block will continue moving at its final speed (which follows from $20~\mathrm{J} = \tfrac12 m v_\mathrm{final}^2$ where m is the mass of the block).

Got it finally. Thank you and dale.

 

[jbriggs444]

What Dale said :-)

To give a numerical example, suppose that I push a block at a constant force of F = 10 N. Initially the block is at rest, and after 2 seconds I stop pushing. The total work on the block is $W = (10~\mathrm N) \cdot (2~\mathrm s) = 20~\mathrm J$. After this there will be no force, so the additional amount of work on the block over any distance s is $W = 0 \cdot s = 0$. The block will continue moving at its final speed (which follows from $20~\mathrm{J} = \tfrac12 m v_\mathrm{final}^2$ where m is the mass of the block).

A force of 10 Newtons multiplied by a time interval of 2 seconds gives an impulse of 20 Newton-seconds = 20 kg m/s

Assuming this force was applied to a 1 kg mass initially at rest, this would produce an acceleration of 10 meters per second and a total displacement (1/2 a t²) of 20 meters during those two seconds. Multiply 10 Newtons by 20 meters and that is 200 Joules of work done.

The final velocity of the 1 kg mass after 2 seconds of acceleration would be 20 meters per second. This is a kinetic energy of 1/2 m v² = 200 Joules. Work done = energy gained.

 

[Imabuleva]

Work-energy theorem:

The work done on a macroscopic system is equal to its change in kinetic energy. Infinite work would require an infinite change in kinetic energy and thus an infinite kinetic energy, which means either an infinite mass or an infinite velocity.

Both possibilities are non-physical, so the answer is no. Work done cannot be infinite.