A load of 60 N is attached to a spring hanging vertically stretches the spring
.050 m. The spring is then put on a horizontal, frictionless table and is
stretched.15 m. What force is required to stretch the spring horizontally for
.15 m?

Question

A load of 60 N is attached to a spring hanging vertically stretches the spring
      .050 m. The spring is then put on a horizontal, frictionless table and is 
      stretched.15 m. What force is required to stretch the spring horizontally for
      .15 m?

anonymous · Answer

F=kx  where F is the force of a spring (Hooke's Law)
you can use the information for the vertical arrangement to find the k (spring constant), when arranged horizontally, the constant remains the same.  So you just need plug in the x to solve for F

anonymous · Answer

but what is k?

anonymous · Answer

you have to solve for it.  every spring has it's own k value.  
F=kx, x is how far it's stretched and F is how much force is applied to stretch it.  Then you solve for k

anonymous · Answer

what do you get for k?

anonymous · Answer

so 60=k(.050) k = 1200
then  F=1200(15)
F=18,000? Did i do this right?

anonymous · Answer

I also thought hookes law had a negative?
F = -kx

anonymous · Answer

the negative sign indicates that this is a restoring force.  Yes, your solution is correct.