cylinder fitted with piston and spring...im lost

[Rival]

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open at the top. Friction is absent. The spring constant of the spring is 2900 N/m. The piston has a negligible mass and a radius of 0.023 m.

(a) When air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress?

(b) How much work does the atmospheric pressure do in compressing the spring?

i am lost...I have no idea as to what equation to use and where to start this problem. any help would be greatly appreciated

Kevin

[turin]

You probably want to post stuff like this in the HW section.

The basic concepts in part a are balance of force, definition of pressure/relationship of pressure to force, and Hooke's law. The initial state has two forces pushing the piston out and one force pushing the pistion in. The final state has one force pushing the piston out and one force pushing the piston in. I won't tell you any more, but I will give you a hint in the form of a question that you should think about: what force is missing in the final state.

The basic concept in part b is work done by a fluid that undergoes a change in volume at constant pressure. Hint: there's probably an equation in the book.

[chroot]

Step 1) To begin with, note that atmospheric pressure is about 14.7 pounds per square inch, or about 101,325 newtons per square meter. Think about what that means: the atmosphere pushes on any object (including your cylinder-piston system) with 101 kilonewtons of force per square meter of surface area.

Step 2) You know the area of the cylinder, so you know how much downward force is applied to it by the atmosphere. Use the "result" of Step 1.

Step 3) You know the spring constant, so you know how much force the spring exerts for a given compression. You know how much force the spring needs to provide to stop the piston from moving -- it needs to push up just as hard as the atmosphere pushes down. Use the result of Step 2.

Step 4) You also know how to calculate the work done by any force acting over a distance:

W = \int F dx

Can you take it from here? Let me know if you don't understand part of this discussion.

- Warren

[Doc Al]

Originally posted by Rival
=i am lost...I have no idea as to what equation to use and where to start this problem. any help would be greatly appreciated

Here's another tip: learn about the properties of springs. How much force they exert and how much energy they store when compressed.

[Rival]

well if i draw a free body diagram of the piston, i will get a force pushing down which would be Atmospheric pressure multiplied by area which is just (pi)(r^2) or in newtons 168.4. now i sitll have a force of 2900 newtons pushing up from the spring, correct? or am i wrong there
kevin

[chroot]

The spring only pushes with 2900 N of force if you compress it one whole meter. If you only compress it half a meter, it only pushes with 1450 N of force.

- Warren

[Rival]

Ah ha!!!

i have F=kx as my equation...F=force pushing down on cylinder and spring

F=168.4N
k=2900N

F/k=x

x=.058m

ok first part done but still stuck on second part

kevin...
P.S. thanks everyone and especially Warren

[chroot]

You're correct, and you're welcome.

For the second part, you need to consider the definition of work:

W = \int F\, dx

The force exerted by the spring (against the atmosphere) is just kx, so the expression becomes

W = \int k x\, dx

Do you know calculus? Can you evaluate this integral?

- Warren

[Doc Al]

Originally posted by Rival
ok first part done but still stuck on second part

Now consider how much work it took to compress the spring. Which is another way of asking how much energy is stored in the compressed spring.

[Rival]

so is it just the kinetic energy in the spring? KE=1/2Mv^2?

Kevin

[chroot]

Spring don't have kinetic energy, unless they're flying across the room. Springs store potential energy. Can you evaluate the integral I posted above that relates W and x for a spring?

- Warren

[Rival]

why yes i can evaluate that integral, or at least i better be able to. seeing how i have finished diff E.Q. last quarter... anyways the equation is just

1/2kx^2

so i have

.5(2900)(.058)^2= 4.8778J

thanks again...definitly a confidence booster right there =)

Kevin

[chroot]

Don't forget your units. (Seriously.) [:D]

- Warren

[Rival]

changes have been made...all units have been placed in the appropriate places for my answers

kevin[;)]