Emergency elevator safety feature calculations

[Achintya]

Homework Statement

The cable of a 4,000 kg elevator snaps when the elevator is at rest on the first floor so that the bottom is a distance d = 12.0 m. A safety device claps the guide rails so that a constant frictional force of 10,000 opposes the motion of the elevator.(g=10 m/s^2)

a) What is the speed of the elevator just before it hits the ground?

Relevant Equations

PLEASE DON'T SOLVE THIS USING CONSERVATION OF ENERGY METHOD...IF POSSIBLE SOLVE THIS USING THE FORCE VELOCITY RELATION(F=vdv/dx)

 

[etotheipi]

Did you forget the mass in $F_y = mv_y\frac{dv_y}{dy}$?

It seems you're slightly overcomplicating it too. It's a constant acceleration question, so you can do it in one line of SUVAT!

Also, you say not to solve it using 'energy' methods, but the relation you've derived from considering forces is (save for that missing factor of $m$) a statement of the work energy theorem. Hopefully you can see that energy approaches and forces approaches are often completely equivalent!

 

[Achintya]

Did you forget the mass in $F_y = mv_y\frac{dv_y}{dy}$?

It seems you're slightly overcomplicating it too. It's a constant acceleration question, so you can do it in one line of SUVAT!

Also, you say not to solve it using 'energy' methods, but the relation you've derived from considering forces is (save for that missing factor of $m$) a statement of the work energy theorem. Hopefully you can see that energy approaches and forces approaches are often equivalent!

thank a lot sir ...i just didn't realize that thing... i know it was silly.

 

[haruspex]

It seems you're slightly overcomplicating it too. It's a constant acceleration question, so you can do it in one line of SUVAT!

There's an instruction to use F=vdv/dx (which, as you say, should be mvdv/dx).