# Physics spring question

1. Mar 5, 2005

### rulers14

The block in Figure 7-11a lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 40 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.8 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops.
Assume that the stopping point is reached.

(a) What is the position of the block?

please someone help me with this problem, its been so annoying. i figured you were to use Hooke's Law to solve it, plugging in 2.8 N for the force and 40 N/m for k. doing this gives you .07. however, the answer is not .07 because it tells me i am wrong when i put that in. does anyone know where i went wrong???

2. Mar 5, 2005

### Staff: Mentor

Find the point where the work done by the applied force equals the spring potential energy. (Note that this is not the equilibrium point, where the net force on the block is zero. It is only a momentary stopping point as the block reverses direction.)

3. Mar 5, 2005

Use the work energy theorem:

$$\Delta W = \Delta E_{Kinetic} + \Delta E_{Spring Potantial} + \Delta E_{Gravity} + \Delta E_{Friction}$$

As Doc Al alwready said, in this case, you only have to deal with:
$$\Delta W = \Delta E_{Spring Potantial}$$

You know the rest.

Regards,

4. Mar 5, 2005

### Andrew Mason

Your answer would be correct if the block had 0 mass. What is the mass of the block?

The block acquires kinetic energy until the applied force = kx (x = F/k), then it loses KE until it is all converted to spring potential energy. It is that KE (at x=F/k) that you need to calculate.

$$ma = F - kx$$

$$KE_{max} = \int_{0}^{F/k}madx$$

See if you can work it out from there.

Edit: You can think of the distance x = F/k as an equilibrium position and the KE of the block at this equilibrium position is eventually converted to spring potential energy at maximum extension (the system oscillates - similar to a block hanging from the spring with mg being the applied force).

AM

Last edited: Mar 5, 2005