# 1kg object and I apply a force of 1N for 1sec, It will move 1/2m

1. Nov 9, 2009

### Mentallic

I'm having trouble understanding work.

Taking an example: If I have a 1kg object and I apply a force of 1N for 1sec, It will move 1/2m and thusI will have done 1/2 (units?) of work, by the equation W=Fs.
But instead, if I consider the same object now travelling at 10m/s and apply a 1N force for 1 sec, the object will travel 10.5m in that time, so now the work done is 10.5?

I don't understand how you can do more work when all you've done is apply the same force to the same object for the same amount of time, but the only difference is that the object in the first case was stationary relative to you, while the other case it was moving. And also, if I applied the force in the direction opposing the motion, less work would have been done than if I applied it in the direction of motion?

2. Nov 9, 2009

### redargon

Re: Work

first of all, how did you work out that the object will only move 0.5m if you apply a 1N force for 1 second to it?

Work is a vector as is force. The work is done by the force in the direction of the force. If an object is already travelling at 10 m/s, the work that you do by applying your 1N for 1 second is the same as in the first example. The force that is causing the object to travel at 10m/s is also doing work of it's own. The work of both together is the total work. Work is just energy and is expressed in Joules [J] usually as this is the SI unit or work and energy.

What's really going to blow your mind is that if you push the object one direction with 1N for 1 second and then push it in the opposite direction with 1N for 1sec and it returns to it's original position, you've done no work. That's the beauty of vectors.

3. Nov 9, 2009

### Mentallic

Re: Work

By using $s=ut+1/2at^2$
$a=F/m$ so therefore a=1 and t=1, u=0.

But there is no force required for the object to move at 10m/s. I'm assuming a non-resistive environment so it will continue at it's usual velocity.

4. Nov 9, 2009

### A.T.

Re: Work

What? From http://en.wikipedia.org/wiki/Work_(physics [Broken]) :

Last edited by a moderator: May 4, 2017
5. Nov 9, 2009

Re: Work

6. Nov 9, 2009

### Jobrag

Re: Work

I think that you are confusing work and power, in the equation for work (force X distance moved) there is no time element, power is work done per unit of time.

7. Nov 9, 2009

### redargon

Re: Work

So the object will have an initital kinetic energy of 1/2mv², so (1/2(m)(10)²) and then you will be increasing that kinetic energy to 1/2m(10.5)² by adding your force in the same direction of the motion and work is the change in kinetic energy. So, W = E2-E1 or 1/2m(10.5²-10²)