Work: an object falls onto a spring. Find max. dist. the spring is compressed.

[DavidAp]

Until I learn to walk my name might pop up a lot in this thread...

A block of mass m = 2.4 kg is dropped from height h = 59 cm onto a spring of spring constant k = 1240 N/m. Find the maximum distance the spring is compressed.

Answer: 0.170m

The first thing I did was convert the height from centimeters to meters so that all the units could work nicely together giving me 0.59 m.

Then I found Work using mgrCos(theta).
W = mgrCos(theta)
W = 2.4 kg(9.8 m/(s^2))(0.59 m)Cos(0)
W = 13.8768 J

After I played around with a more general equation of Work.
W = ΔK + ΔU
W = Kf - Ki + Uf - Ui

However, since the spring is in equilibrium before the block falls on it, conceptually, I saw that the Ui should equal to zero so my equation is,
W = Kf - Ki +Uf

And we know that Ki = 1240 N/m so,
W = Kf - 1240 N/m + Uf
W = 1/2m(v^2) - 1240 N/m + mgy

13.8768 J = 1/2(2.4 kg)(Vf^2) - 1240 N/m + 2.4 kg(9.8 m/(s^2))(y)
Now, I wrote Vf because it's an unknown and though I think there is a way to solve it using Work I went on to use a kinematic equation because I'm still very new to Work and have not yet learned how to do that.
Vf^2 = Vi^2 +2aΔy
Vf^2 = 9.8 m/(s^2)(0.59 m)
Vf^2 = 11.564 m/s

Back to my equation,
13.8768 J = 1/2(2.4 kg)(11.564 m/2) - 1240 N/m + 23.52 N(y)
13.8768 J = 13.8768 J - 1240 N/m + 23.52 N(y)
0 = -1240 N/m + 23.52 N(y)
1240 N/m = 23.52 N(y)
52.72 /m = y
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As you can see something went horribly wrong! I'm new to Work but I felt pretty confident approaching this problem until I got to the end. Not only is my answer numerically incorrect but the units seem impossible; nothing over meters?

Can somebody help show me what I did wrong with this question? I'm sorry if it was difficult to follow, mathematics on text always seems like a puzzle to me.

Thank you for taking the time to review my question, I greatly appreciate it.

 

[bjd40@hotmail.com]

i think u r wrong at taking the height that the object falls as 0.59m. actually the height should be 0.59 +x, where x is the compression of the spring. at that point there is no KE. so, the relevant eqn. should be (from conservation of energy),
mg(h+x) = 1/2 kx2.
this is a quadratic eqn. in x. now solve for x, if u know how to solve a quadratic eqn.

 

[DavidAp]

i think u r wrong at taking the height that the object falls as 0.59m. actually the height should be 0.59 +x, where x is the compression of the spring. at that point there is no KE. so, the relevant eqn. should be (from conservation of energy),
mg(h+x) = 1/2 kx2.
this is a quadratic eqn. in x. now solve for x, if u know how to solve a quadratic eqn.

Thank you, you are absolutely right! The math is so much shorter this way too!