Position of Block When Spring Stops Pulling

[mochabcha]

position of a spring!

Okay, Here's the problem
The block in Figure 7-11a (Figure not important) lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 35 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? (b) What is the work that has been done on the block by the applied force? (c) What is the work that has been done on the block by the spring force?

(a) I drew a free body diagram and it seems like when the block stops the spring force catches up to the applied force counteracting it and stopping the motion of the block. So, Hooke's law states that F(spring) = -kx; since F(spring) = F(applied) --> F(applied) = -kx.
I crunched those numbers and got -.08 which made sense because the applied force is going to be positive [f(s) is negative in Hooke's law and x being a negative number will cancel the negatives].
BUT THAT'S WRONG what's the problem!

 

[Electro]

Hello,
I have been having troubles with these springs too but I will try to explain where the problem stands with your exercise.
Since the force pulls i.e. is directed +x-axis, it means that the force is negative because the force of spring has to balance the applied force. So it is -2.8N. Thus u can use the equation and end up with 0.08 m. Then the work should be negative (W = - .5(k x^2)
That's all I think

P.S: By the way, are you using the book which has a lightning in the front cover?

 

[greenman100]

Hello,
I have been having troubles with these springs too but I will try to explain where the problem stands with your exercise.
Since the force pulls i.e. is directed +x-axis, it means that the force is negative because the force of spring has to balance the applied force. So it is -2.8N. Thus u can use the equation and end up with 0.08 m. Then the work should be negative (W = - .5(k x^2)
That's all I think

P.S: By the way, are you using the book which has a lightning in the front cover?

x=2*force/k

 

[chroot]

In a problem as simple as this, you don't really even need to get caught up in the whole negative/positive controversy. In fact, it's generally easier in physics to just think about the sign your final answer should have, rather than fussing with keeping the signs consistent the whole way through your calculation.

The block is being pulled to the right via some externally supplied force of 2.8N. The restoring force due to the spring pulls the block back to the left. When the block stops, the forces must be the same in both directions.

Don't worry about the signs. Just figure out how far you'd have to stretch the spring to get it to generate a force with a magnitude of 2.8N. Since the block was originally pulled in the direction of the positive x-axis, it's obvious that the block will be on the positive x-axis, and thus the distance will be positive.

- Warren