Spring force with free fall problem

[scarne92]

Homework Statement

Consider the system below. The 2.50 kg mass compresses the spring (k = 805 N/m) a distance of 0.230 m from equilibrium. The mass is then released from rest. It slides a total distance of 1.20 m on the table top where it feels a force of kinetic friction of 5.60 N before it slides off the edge. If it hits the ground with a speed of 6.64 m/s, find the height ∆y of the table.

Homework Equations

Vf^2 = Vi^2-2g∆y

The Attempt at a Solution

I started doing the spring force part of the problem, but then the thought hit me that the speed shouldn't matter in the x direction because it is just a free fall problem. I'm not sure if I'm correct in that thinking though.

here's what I got.

∆y = (Vf^2 - Vi^2)/-2g

∆y = (6.64^2 - 0^2)/-2(9.81)

∆y = 44.0896/-19.62

∆y = 2.25m

 

[PeterO]

Homework Statement

Consider the system below. The 2.50 kg mass compresses the spring (k = 805 N/m) a distance of 0.230 m from equilibrium. The mass is then released from rest. It slides a total distance of 1.20 m on the table top where it feels a force of kinetic friction of 5.60 N before it slides off the edge. If it hits the ground with a speed of 6.64 m/s, find the height ∆y of the table.

Homework Equations

Vf^2 = Vi^2-2g∆y

The Attempt at a Solution

I started doing the spring force part of the problem, but then the thought hit me that the speed shouldn't matter in the x direction because it is just a free fall problem. I'm not sure if I'm correct in that thinking though.

here's what I got.

∆y = (Vf^2 - Vi^2)/-2g

∆y = (6.64^2 - 0^2)/-2(9.81)

∆y = 44.0896/-19.62

∆y = 2.25m

It is not a simple free fall problem.

If the mass just tipped over the edge, it just falls to the floor reaching a speed of 6.64

If however the mass leaves the table traveling at a speed of 6.6 m/s, it will have gained the extra 0.04 m/s by falling just a few mm.

In the end this is a projectile motion question: you have to work out how fast the mass is traveling when it leaves the table.