Ugh, I don't understand. Consider four masses, A-D, attached to springs as shown. All springs are identical, with spring constant k, and are in their natural, unstretched state. Let FA(d) denote the force that must be applied to displace mass A a distance d from equilibrium. Let WA(d) denote the work done on mass A by the spring forces when it is moved from equilibrium to position d.

http://i42.tinypic.com/9s4cnp.gif

What is WB(d)/WD(2d)

Can someone explain this to me? Really appreciate it

Question

Ugh, I don't understand. Consider four masses, A-D, attached to springs as shown. All springs are identical, with spring constant k, and are in their natural, unstretched state. Let FA(d) denote the force that must be applied to displace mass A a distance d from equilibrium. Let WA(d) denote the work done on mass A by the spring forces when it is moved from equilibrium to position d. 

http://i42.tinypic.com/9s4cnp.gif

What is WB(d)/WD(2d) 

Can someone explain this to me? Really appreciate it

anonymous · Answer

I also have a question like:

 FB(d)/FC(d)  ?

For work, I know we can use F*d

anonymous · Answer

It seems easy but I dont know where to start

anonymous · Answer

Take the first two for an example.  What would be more work, FA(d) or FB(d)?

anonymous · Answer

FB(d) but how do i get a numerical answer?

anonymous · Answer

anyone?

anonymous · Answer

The idea is, for example, if you add two springs next to one another you've effectively doubled the spring constant, or stiffness.  Therefore, you'd be doing twice as much work as you would with just one.

On the other hand, if you connect two springs end to end, then to move the mass a distance d you'd only need each spring to stretch by a distance d/2, which means you'd be doing half as much work as you would with a single spring.