# Work done by or on a system?

1. Aug 28, 2014

### daigiaga1994

Consider a system with a spring and a block. The block moves on a frictionless surface. At the equilibrium position, the speed of the block equals to 0. Now i pull the block by a force F with a displacement of x. Assuming that the speed at the final position is v.
Can you calculate the energy of the system after the process ?

If i apply the equation: ΔE= Q+W1 ( with W1 is the work donr on the system )
Is the above equation equivalent with the eqution: ΔE=Q-W2( with W2 is the work done by the system)

( Q is the heat that is transfered to the system)
If these equations is equivalent, please tell me how to calculate W2 ?
thank

2. Aug 28, 2014

### Staff: Mentor

Is the force constant over that displacement?

(The work done by the system is just the negative of the work done on the system.)

3. Aug 28, 2014

### daigiaga1994

I think it is not important if F is constant or not. Because we can represent the work done on the system by W1=∫Fdx.

In my opinion, W2 is the work done by the system, so it is done by the spring force ( This is only force created by the system) and equals to W2=-1/2kx2.

If the two equation is equivalent, W1+W2=0. It is clearly wrong because W1+W2=ΔK=1/2mv2.

This is my confusion

4. Aug 29, 2014

### Staff: Mentor

This is where you are going wrong. If you pull the block with a force F, then the block is pulling you with a force -F. (Newton's 3rd law.) So the work done by the system W2= -∫Fdx.

5. Aug 29, 2014

### daigiaga1994

Thank you so much Doc Al. Now i know what is the work done by the system.