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 _______________________________________________ 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.